If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

# Past Exam Questions, Solutions, and Explanations on Circle Theorems

Pre-requisites:
(1.) Expressions and Equations
(2.) Plane Geometry: Points, Lines, and Angles
(3.) Triangles
(4.) Length of Arc

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Questions with multiple choice options are for the WASSCE General Mathematics Paper 1 (Objective)
Questions without multiple choice options are for the WASSCE General Mathematics Paper 2 (Theory)

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
State the reason(s) for each step.
Show all work.

Some of these questions can be solved in at least two (two or more) ways.
Teachers can encourage students to get the same solution using more than one approach/method.

Where applicable, depending on time; I solved some questions using more than one approach.

(1.) CSEC The diagram below shows a circle where $AC$ is a diameter.
$B$ and $D$ are two other points on the circle and $DCE$ is a straight line.
Angle $CAB = 28^\circ$ and $\angle DBC = 46^\circ$
Calculate the value of each of the following angles.
Show detailed working where necessary and give a reason to support your answers.

$(i)\:\: \angle DBA \\[3ex] (ii)\:\: \angle DAC \\[3ex] (iii)\:\: \angle BCE \\[3ex]$

$(i) \\[3ex] \angle ABC = 90^\circ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle ABC = \angle DBA + \angle DBC ...diagram \\[3ex] 90 = \angle DBA + 46 \\[3ex] \angle DBA + 46 = 90 \\[3ex] \angle DBA = 90 - 46 \\[3ex] \angle DBA = 44^\circ \\[3ex] (ii) \\[3ex] \angle DAC = \angle DBC = 46^\circ ... \angle s \:\:in\:\:the\:\:same\:\:segment\;\;are\;\;equal \\[3ex] (iii) \\[3ex] \angle BCE = \angle A ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] \angle A = \angle DAC + \angle CAB ...diagram \\[3ex] \angle A = 46 + 28 \\[3ex] \angle A = 74 \\[3ex] \therefore \angle BCE = 74^\circ$
(2.) curriculum.gov.mt The figure below shows triangle ABC inscribed in a circle centre O.
BO and AO are radii drawn to form triangle ABO.

(b) What type of triangle is ABO? Explain.
(c) Work out angle ABO.

$(a) \\[3ex] \angle AOB = 2 * \angle ACB ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOB = 2(40) \\[3ex] \angle AOB = 80^\circ \\[3ex] (b) \\[3ex] \triangle ABO \;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] because\;\; |OA| = |OB| \\[3ex] ... same\;\;radius\;\;of\;\;the\;\;circle = same\;\;side\;\;of\;\;\triangle \\[3ex] and \\[3ex] because\;\;\angle OAB = \angle OBA ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] (c) \\[3ex] \angle ABO = \angle BAO = p ..base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \\[3ex] \angle ABO + \angle BAO + \angle AOB = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] p + p + 80 = 180 \\[3ex] 2p = 180 - 80 \\[3ex] 2p = 100 \\[3ex] p = \dfrac{100}{2} \\[5ex] p = 50 \\[3ex] \therefore \angle ABO = \angle BAO = 50^\circ$
(3.) WASSCE
The diagram is a circle with centre $O$
$U, V, W$ and $Y$ are points on the circle.
Find $\angle UVW$

$A.\:\: 144^\circ \\[3ex] B.\:\: 72^\circ \\[3ex] C.\:\: 40^\circ \\[3ex] D.\:\: 36^\circ \\[3ex]$

$Obtuse\;\angle UOW = 2(x) = 2x ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 8x + 2x = 360 ...\angle s\:\:at\:\:a\:\:point \\[3ex] 10x = 360 \\[3ex] x = \dfrac{360}{10} \\[5ex] x = 36 \\[3ex] Reflex\;8x = 2 * \angle UVW ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 8x = 8 * 36 = 288 \\[3ex] \implies \\[3ex] 288 = 2 * \angle UVW \\[3ex] \angle UVW = \dfrac{288}{2} \\[5ex] \angle UVW = 144^\circ \\[3ex] OR \\[3ex] \angle UVW + \angle UYW = 180^\circ ...interior\:\:opposite\;\;\angle s\:\:of\:\:a\:\:cyclic\:\:quadrilateral\;\;are\;\;supplementary \\[3ex] \angle UYW = x = 36 \\[3ex] \angle UVW + 36 = 180 \\[3ex] \angle UVW = 180 - 36 \\[3ex] \angle UVW = 144^\circ$
(4.) JAMB In the figure below, find $PRQ$
$A.\:\: 66\dfrac{1}{2}^\circ \\[5ex] B.\:\: 62\dfrac{1}{2}^\circ \\[5ex] C.\:\: 125^\circ \\[3ex] D.\:\: 105^\circ \\[3ex] E.\:\: 65^\circ \\[3ex]$

$\angle O = 235 + y = 360 ...\angle s \:\:at\:\:a\:\:point \\[3ex] y = 360 - 235 \\[3ex] y = 125 \\[3ex] y = 2x ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] \implies \\[3ex] 125 = 2x \\[3ex] 2x = 125 \\[3ex] x = \dfrac{125}{2} \\[5ex] x = 62\dfrac{1}{2}^\circ$
(5.) WASSCE
In the diagram, $O$ is the centre of the circle, $P\hat{O}R = 108^\circ$ and $\angle ORK = 20^\circ$
Calculate $\angle OPK$.

$A.\:\: 27^\circ \\[3ex] B.\:\: 34^\circ \\[3ex] C.\:\: 36^\circ \\[3ex] D.\:\: 42^\circ \\[3ex]$

Construction: Join the chord from point P to point R
$\angle OPR = \angle ORP = y ...base\:\:\angle s \:\:of\:\:isosceles\:\: \triangle OPR \\[3ex] \underline{\triangle OPR} \\[3ex] \angle OPR + \angle ORP + \angle POR = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle OPR \\[3ex] y + y + 108 = 180 \\[3ex] 2y = 180 - 108 \\[3ex] 2y = 72 \\[3ex] y = \dfrac{72}{2} \\[5ex] y = 36 \\[3ex] \therefore \angle OPR = \angle ORP = 36^\circ \\[3ex] \angle POR = 2 * \angle PKR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PKR \\[3ex] \angle PKR = \dfrac{108}{2} \\[5ex] \angle PKR = 54 \\[3ex] \underline{\triangle PKR} \\[3ex] \angle KPR + \angle KRP + \angle PKR = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle PKR \\[3ex] \angle KPR = \angle OPK + \angle OPR...diagram \\[3ex] \angle KPR = \angle OPK + 36 \\[3ex] \angle KRP = 20 + \angle ORP...diagram \\[3ex] \angle KRP = 20 + 36 \\[3ex] \angle KRP = 56^\circ \\[3ex] \implies \\[3ex] \angle OPK + 36 + 56 + 54 = 180 \\[3ex] \angle OPK + 146 = 180 \\[3ex] \angle OPK = 180 - 146 \\[3ex] \angle OPK = 34^\circ$
(6.) GCSE This circle has centre C.
W, X and Y are points on the circle.
WY is a straight line.

Tick one box for each statement.

WY is a diameter ... True
This is because it is a chord that passes through the centre of the circle.

WX is a radius. ... False
It does not pass through the centre of the circle.

The shaded section is a sector. ... False
It is a segment. It is the minor segment.
A sector will also include the area enclosed by both radii: CW and CX
In other words, the minor sector will be the area of ▵WCX and the area of minor segment WX

Arc XY is part of the circumference. ... True
Arcs of a circle are part of the circumference of the circle. $(7.) CSEC The diagram below, not drawn to scale, shows a circle with centre$O$.$HJ$and$HG$are tangents to the circle and$\angle JHG = 48^\circ$Calculate, giving the reason for each step of your answer, the measure of:$ (i)\:\: \angle OJH \\[3ex] (ii)\:\: \angle JOG \\[3ex] (iii)\:\: \angle JKG \\[3ex] (iv)\:\: \angle JLG \\[3ex]  (i) \\[3ex] \angle OJH = 90^\circ ... radius\;OJ \perp tangent\;JH\:\:at\:\:point\:\:of\:\:contact\;J \\[3ex] (ii) \\[3ex] \angle OGH = 90^\circ ... radius\;OG \perp tangent\;GH\:\:at\:\:point\:\:of\:\:contact\;G \\[3ex] \underline{Quadrilateral\:\:JOGH} \\[3ex] \angle J + \angle O + \angle G + \angle H = 360 ...sum\:\:of\:\:interior\:\:\angle s\:\:of\:\:a\:\:Quadrilateral \\[3ex] 90 + \angle O + 90 + 48 = 360 \\[3ex] 228 + \angle O = 360 \\[3ex] \angle O = 360 - 228 \\[3ex] \angle O = 132 \\[3ex] \angle O = \angle JOG = 132^\circ \\[3ex] (iii) \\[3ex] Obtuse\;\;\angle JOG = 2 * \angle JKG ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 132 = 2 * \angle JKG \\[3ex] 2 * \angle JKG = 132 \\[3ex] \angle JKG = \dfrac{132}{2} \\[5ex] \angle JKG = 66^\circ \\[3ex] (iv) \\[3ex] Reflex\;\;\angle JOG + Obtuse\;\;\angle JOG = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle JOG + 132 = 360 \\[3ex] Reflex\;\;\angle JOG = 360 - 132 \\[3ex] Reflex\;\;\angle JOG = 228 \\[3ex] Reflex\;\;\angle JOG = 2 * \angle JLG ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 228 = 2 * \angle JLG \\[3ex] 2 * \angle JLG = 228 \\[3ex] \angle JLG = \dfrac{228}{2} \\[5ex] \angle JLG = 114^\circ \\[3ex] OR \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:KJLG} \\[3ex] \angle K + \angle L = 180^\circ ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral \\[3ex] 66 + \angle L = 180 \\[3ex] \angle L = 180 - 66 \\[3ex] \angle L = 114 \\[3ex] \angle L = \angle JLG = 114^\circ $(8.) WASSCE In the diagram,$O$is the center of the circle$PQRS$and$\angle PQR = 65^\circ$. Find the value of the angle$x A.\:\: 115^\circ \\[3ex] B.\:\: 130^\circ \\[3ex] C.\:\: 230^\circ \\[3ex] D.\:\: 255^\circ \\[3ex]  Obtuse\;\;\angle POR = 2(65) ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] Obtuse\;\;\angle POR = 130^\circ \\[3ex] x + Obtuse\;\;\angle POR = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] x + 130 = 360 \\[3ex] x = 360 - 130 \\[3ex] x = 230^\circ $(9.) WASSCE In the diagram,$O$is the centre of the circle$PQS$.$\overline{SR}$is a tangent, reflex$P\hat{O}S = 252^\circ$and$\angle SQR = 79^\circ$. Calculate the size of$\angle QRS Reflex\:\:P\hat{O}S = 252^\circ \\[3ex] \angle SQR = 79^\circ \\[3ex] \angle QRS = x \\[3ex] Reflex\:\:P\hat{O}S + Obtuse\:\:P\hat{O}S = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] 252 + Obtuse\:\:P\hat{O}S = 360 \\[3ex] Obtuse\:\:P\hat{O}S = 360 - 252 \\[3ex] Obtuse\:\:P\hat{O}S = 108^\circ \\[3ex] Obtuse\:\:P\hat{O}S = 2 * \angle PQS ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PQS \\[3ex] \angle PQS = \dfrac{108}{2} \\[5ex] \angle PQS = 54^\circ \\[3ex] $Construction: Join the chord from point P to point S$ \underline{\triangle PQS} \\[3ex] \angle QPS = \angle QSP = y...base\:\:\angle s\:\:of\:\:isosceles\:\:\triangle PQS \\[3ex] \angle QPS + \angle QSP + \angle PQS = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle PQS \\[3ex] \implies \\[3ex] y + y + 54 = 180 \\[3ex] 2y = 180 - 54 \\[3ex] 2y = 126 \\[3ex] y = \dfrac{126}{2} \\[5ex] y = 63 \\[3ex] \therefore \angle QPS = \angle QSP = 63^\circ \\[3ex] \angle QSR = \angle QPS = 63^\circ ...\angle\;\;between\;\;tangent\;SR\;\;and\;\;chord\;SQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle QRS} \\[3ex] \angle QSR + \angle SQR + \angle QRS = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle QRS \\[3ex] \implies \\[3ex] 63 + 79 + \angle QRS = 180 \\[3ex] 142 + \angle QRS = 180 \\[3ex] \angle QRS = 180 - 142 \\[3ex] \angle QRS = 38^\circ $(10.) GCSE The diagram shows a circle with centre O. P, Q and R are points on the circle. Choose words from the box to complete these sentences. (i) Line OP is a ........................ (ii) Line QR is a ........................ (iii) Lines OP and QR are ........................ (i) Line OP is a radius (ii) Line QR is a chord (iii) Lines OP and QR are parallel because of the similar single arrows (11.) JAMB$PQRS$is a cyclic quadrilateral in which$PQ = PS$.$PT$is a tangent to the circle and$PQ$makes an angle of$50^\circ$with the tangent as shown in the figure below. What is the size of$QRS$?$ A.\:\: 50^\circ \\[3ex] B.\:\: 40^\circ \\[3ex] C.\:\: 110^\circ \\[3ex] D.\:\: 80^\circ \\[3ex] E.\:\: 100^\circ \\[3ex]  \underline{\triangle PSQ} \\[3ex] \angle PSQ = 50^\circ ...\angle \:\:between\:\:tangent\;PT\:\:and\:\:chord\;PQ = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle PSQ = \angle PQS = 50 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle PSQ \\[3ex] \angle PSQ + \angle PQS + \angle SPQ = 180 ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle PSQ \\[3ex] 50 + 50 + \angle SPQ = 180 \\[3ex] 100 + \angle SPQ = 180 \\[3ex] \angle SPQ = 180 - 100 \\[3ex] \angle SPQ = 80 \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:SPQR} \\[3ex] \angle P + \angle R = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] 80 + \angle R = 180 \\[3ex] \angle R = 180 - 80 \\[3ex] \angle R = \angle QRS = 100^\circ $(12.) WASSCE In the diagram,$O$is the centre of the circle.$\overline{OC}$is parallel to$\overline{AD}$,$\overline{AB}$is a straight line and$\angle OCA = 48^\circ$. Calculate$\angle ABC \angle OCA = \angle OAC = 48^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle OAC \\[3ex] \angle ACB = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle ABC \\[3ex] \angle CAB = \angle OAC = 48^\circ ...diagram \\[3ex] \implies \\[3ex] 48 + \angle ABC + 90 = 180 \\[3ex] 138 + \angle ABC = 180 \\[3ex] \angle ABC = 180 - 138 \\[3ex] \angle ABC = 42^\circ $(13.) CSEC The diagram below, not drawn to scale, shows a circle with centre$O$. The vertices$H, J, K\:\:and\:\:L$of a quadrilateral lie on the circumference of the circle and$PKM$is a tangent to the circle at$K$. The measure of angle$H\hat{J}L = 20^\circ$and$J\hat{K}H = 50^\circ$Calculate, giving reasons for each step of your answer, the measure of$ (i)\:\: H\hat{K}L \\[3ex] (ii)\:\: J\hat{O}K \\[3ex] (iii)\:\: J\hat{H}K \\[3ex]  (i) \\[3ex] H\hat{K}L = \angle HJL = 20^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \underline{\triangle JOK} \\[3ex] \angle OKJ = \angle OJK = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle JOK \\[3ex] \angle OKJ + \angle OJK + \angle JOK = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle JOK \\[3ex] 50 + 50 + \angle JOK = 180 \\[3ex] 100 + \angle JOK = 180 \\[3ex] \angle JOK = 180 - 100 \\[3ex] \angle JOK = 80^\circ \\[3ex] (iii) \\[3ex] \angle HJK = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle HJK} \\[3ex] \angle JHK + \angle HJK + \angle JKH = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle HJK \\[3ex] \angle JKH + 90 + 50 = 180 \\[3ex] \angle JHK + 140 = 180 \\[3ex] \angle JHK = 180 - 140 \\[3ex] \angle JHK = 40^\circ \\[5ex] OR \\[3ex] Another\;\;Approach\;\;to\;\;solving\;\;the\;\;question\;\;(longer\;\;approach) \\[3ex] (i) \\[3ex] \angle HJK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle HJK = \angle HJL + \angle LJK...diagram \\[3ex] 90 = 20 + \angle LJK \\[3ex] \angle LJK = 90 - 20 \\[3ex] \angle LJK = 70 \\[3ex] \angle LHK = \angle LJK ...\angle s \:\:on\:\:a\:\:straight\:\:line \\[3ex] \rightarrow \angle LHK = 70 \\[3ex] \angle HLK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle HKL} \\[3ex] \angle LHK + \angle HLK + \angle HKL = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 70 + 90 + \angle HKL = 180 \\[3ex] 160 + \angle HKL = 180 \\[3ex] \angle HKL = 180 - 160 \\[3ex] \angle HKL = 20^\circ \\[3ex] (ii) \\[3ex] \angle OKJ = \angle OJK ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \rightarrow \angle OJK = 50 \\[3ex] \underline{\triangle JOK} \\[3ex] \angle OKJ + \angle OJK + \angle JOK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 50 + 50 + \angle JOK = 180 \\[3ex] 100 + \angle JOK = 180 \\[3ex] \angle JOK = 180 - 100 \\[3ex] \angle JOK = 80^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle JHK} \\[3ex] \angle HJK + \angle JKH + \angle JHK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 90 + 50 + \angle JHK = 180 \\[3ex] 140 + \angle JHK = 180 \\[3ex] \angle JHK = 180 - 140 \\[3ex] \angle JHK = 40^\circ $(14.) WASSCE Find the value of$\angle TSP$in the diagram.$ A.\:\: 91^\circ \\[3ex] B.\:\: 89^\circ \\[3ex] C.\:\: 71^\circ \\[3ex] D.\:\: 69^\circ \\[3ex]  \angle TQP = \angle QTR + \angle QRT \\[3ex] ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle TQP = 31 + 58 \\[3ex] \angle TQP = 89^\circ \\[3ex] \angle TSP = \angle TQP ... \angle s \:\:in\:\:the\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \therefore \angle TSP = 89^\circ $(15.) WASSCE In the diagram,$\angle ZXY = y^\circ$,$\angle ZYX = (y + 17)^\circ$,$\angle PZX = x^\circ$, and$\angle YZT = (2x - 43)^\circ$Find the value of y.$ A.\:\: 40^\circ \\[3ex] B.\:\: 36^\circ \\[3ex] C.\:\: 20^\circ \\[3ex] D.\:\: 9^\circ \\[3ex]  x = y + 17...eqn.(1) ...\angle \:\:between\:\:tangent\:PZT\;\:and\:\:chord\;XZ = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] 2x - 43 = y...eqn.(2) ...\angle \:\:between\:\:tangent\:PZT\:\;and\:\:chord\;YZ = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] Substitute\:\:eqn.(1)\:\:into\:\:eqn.(2) \\[3ex] 2(y + 17) - 43 = y \\[3ex] 2y + 34 - 43 = y \\[3ex] 2y - y = 43 - 34 \\[3ex] y = 9^\circ $(16.) JAMB$PQ$and$PR$are tangents from$P$to a circle centre$O$as shown in the figure above. If$\angle QRP = 34^\circ$, find the angle marked$x A.\:\: 34^\circ \\[3ex] B.\:\: 56^\circ \\[3ex] C.\:\: 68^\circ \\[3ex] D.\:\: 112^\circ \\[3ex]  \angle ORP = 90^\circ \\[3ex] ... radius\;OR \perp tangent\;RP\:\:at\:\:point\:\:of\:\:contact\;R \\[3ex] \angle ORP = \angle ORQ + \angle QRP...diagram \\[3ex] 90 = \angle ORQ + 34 \\[3ex] \angle ORQ = 90 - 34 \\[3ex] \angle ORQ = 56^\circ \\[3ex] \underline{\triangle ROQ} \\[3ex] \angle ORQ = \angle OQR = 56^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle ROQ \\[3ex] \angle ORG + \angle OQR + \angle ROQ = 180 ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle ORG \\[3ex] \implies \\[3ex] 56 + 56 + x = 180 \\[3ex] 112 + x = 180 \\[3ex] x = 180 - 112 \\[3ex] x = 68^\circ $(17.) CSEC The diagram below, not drawn to scale, shows a circle with centre$O$. The points$A, B, C\:\:and\:\:D$are on the circumference of the circle.$EAF$and$EDG$are tangents to the circle at$A$and$D$respectively.$A\hat{O}D = 114^\circ$and$C\hat{D}G = 18^\circ$Calculate, giving reasons for EACH step of your answer, the measure of$ (i)\:\: A\hat{C}D \\[3ex] (ii)\:\: A\hat{E}D \\[3ex] (iii)\:\: O\hat{A}C \\[3ex] (iv)\:\: A\hat{B}C \\[3ex] $We can solve (i) in at least two ways$ (i) \\[3ex] \underline{First\:\:Method} \\[3ex] \angle AOD = 2 * \angle ACD ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 114 = 2 * \angle ACD \\[3ex] \angle ACD = \dfrac{114}{2} \\[5ex] \angle ACD = 57^\circ \\[3ex] \underline{Second\:\:Method} \\[3ex] \angle ODG = 90 ... radius\;OD \perp tangent\;EDG\:\:at\:\:point\:\:of\:\:contact\;D \\[3ex] \angle ODA = \angle OAD = y ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OAD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle OAD \\[3ex] y + y + 114 = 180 \\[3ex] 2y + 114 = 180 \\[3ex] 2y = 180 - 114 \\[3ex] 2y = 66 \\[3ex] y = \dfrac{66}{2} \\[5ex] y = 33 \\[3ex] \angle ADE + \angle ODA + \angle ODG = 180^\circ ...\angle s\:\:on\:\:a\:\:straight\:\:line \\[3ex] \angle ADE + 33 + 90 = 180 \\[3ex] \angle ADE + 123 = 180 \\[3ex] \angle ADE = 180 - 123 \\[3ex] \angle ADE = 57 \\[3ex] \angle ADE = \angle ACD = 57^\circ ...\angle \:\:between\:\:tangent\;EDG\:\:and\:\:chord\;AD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] (ii) \\[3ex] \angle EAD = \angle ACD = 57^\circ ...\angle \:\:between\:\:tangent\;EAF\:\:and\:\:chord\;AD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] \underline{\triangle AED} \\[3ex] \angle ADE + \angle EAD + \angle AED = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle AED \\[3ex] 57 + 57 + \angle AED = 180 \\[3ex] 114 + \angle AED = 180 \\[3ex] \angle AED = 180 - 114 \\[3ex] \angle AED = 66^\circ \\[3ex] (iii) \\[3ex] \angle OAD = \angle OAC + \angle CAD ...diagram \\[3ex] \angle CDG = \angle CAD = 18^\circ ...\angle \:\:between\:\:tangent\;EDG\:\:and\:\:chord\;CD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] \implies \\[3ex] 33 = \angle OAC + 18 \\[3ex] \angle OAC = 33 - 18 \\[3ex] \angle OAC = 15^\circ \\[3ex] (iv) \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:ABCD} \\[3ex] \angle B + \angle D = 180^\circ ...sum\:\:of\:\:the\;\;two\;\;interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:quadrilateral \\[3ex] \angle D = \angle ODA + \angle ODC ...diagram \\[3ex] \angle ODG = \angle ODC + \angle CDG ...diagram \\[3ex] 90 = \angle ODC + 18 \\[3ex] \angle ODC = 90 - 18 \\[3ex] \angle ODC = 72 \\[3ex] So:\;\; \angle D = 33 + 72 \\[3ex] \angle D = 105^\circ \\[3ex] \implies \\[3ex] \angle B + 105 = 180 \\[3ex] \angle B = 180 - 105 \\[3ex] \angle B = 75^\circ \\[3ex] A\hat{B}C = \angle B = 75^\circ $(18.) JAMB In the diagram above,$O$is the center of the circle.$POM$is a diameter and$\angle MNQ = 42^\circ$Calculate$\angle QMP A.\:\: 138^\circ \\[3ex] B.\:\: 132^\circ \\[3ex] C.\:\: 42^\circ \\[3ex] D.\:\: 48^\circ \\[3ex]  \angle MQP = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle MPQ = \angle MNQ = 42^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \underline{\triangle QMP} \\[3ex] \angle QMP + \angle MPQ + \angle MQP = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle QMP \\[3ex] \angle QMP + 42 + 90 = 180 \\[3ex] \angle QMP + 132 = 180 \\[3ex] \angle QMP = 180 - 132 \\[3ex] \angle QMP = 48^\circ $(19.) WASSCE In the diagram,$O$is the center of the circle.$\angle OQR = 32^\circ$and$\angle TPQ = 15^\circ$Calculate:$ (i)\:\: \angle QPR \\[3ex] (ii)\:\: \angle TQO \\[3ex]  (i) \\[3ex] \angle OQR = \angle ORQ = 32 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OQR} \\[3ex] \angle OQR + \angle ORQ + \angle QOR = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle OQR \\[3ex] 32 + 32 + \angle QOR = 180 \\[3ex] 64 + \angle QOR = 180 \\[3ex] \angle QOR = 180 - 64 \\[3ex] \angle QOR = 116 \\[3ex] \angle QOR = 2 * \angle QPR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 116 = 2 * \angle QPR \\[3ex] \angle QPR = \dfrac{116}{2} \\[5ex] \angle QPR = 58^\circ \\[3ex]  (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:QTPR} \\[3ex] \angle P + \angle Q = 180^\circ ...sum\:\:of\:\:the\;\;two\;\;interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:quadrilateral \\[3ex] \angle P = \angle TPQ + \angle QPR ...diagram \\[3ex] \angle P = 15 + 58 \\[3ex] \angle P = 73^\circ \\[3ex] \angle Q = \angle TQO + \angle OQR ...diagram \\[3ex] \angle Q = \angle TQO + 32 \\[3ex] \implies \\[3ex] 73 + \angle TQO + 32 = 180 \\[3ex] \angle TQO + 105 = 180 \\[3ex] \angle TQO = 180 - 105 \\[3ex] \angle TQO = 75^\circ $(20.) curriculum.gov.mt YZ is a diameter of the circle. What type of triangle is XYZ?$ \angle YXZ = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \\[3ex] \triangle XYZ \;\;is\;\;a\;\;right-angled\;\;\triangle $(21.) CSEC (a) In the diagram below,$VWZ$and$WXYZ$are two circles intersecting at$W$and$Z$.$SVT$is a tangent to the circle at$VVWX$and$VZY$are straight lines.$T\hat{V}Y = 78^\circ$and$S\hat{V}X = 51^\circ$(i) Calculate the size of EACH of the following angles, giving reasons for your answers.$ a)\:\: V\hat{Z}W \\[3ex] b)\:\: X\hat{Y}Z \\[3ex]  a) \\[3ex] \angle SVW = \angle VZW = 51^\circ ...\angle \:\:between\:\:tangent\;SVT\:\:and\:\:chord\;VW = \angle\:\:in\;\;the\:\:alternate\:\:segment \\[3ex] b) \\[3ex] \angle TVZ = \angle VWZ = 78^\circ ...\angle \:\:between\:\:tangent\;SVT\:\:and\:\:chord\;VZ = \angle\:\:in\;\;the\:\:alternate\:\:segment \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:WXYZ} \\[3ex] \angle VWZ = \angle XYZ = 78^\circ ...exterior\:\: \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\:\:opposite\:\: \angle $(22.) ACT The circle below has its center at O, and points A and B lie on the circle. The length of$\overline{AO}$is 4 inches, and arc$\overset{\huge\frown}{AB}$measures 112° What is the measure of ∠ABO?$ F.\;\; 17^\circ \\[3ex] G.\;\; 34^\circ \\[3ex] H.\;\; 45^\circ \\[3ex] J.\;\; 56^\circ \\[3ex] K.\;\; 68^\circ \\[3ex] $Intercepted arc = angle at center (central angle)$ \angle AOB = \overset{\huge\frown}{AB} = 112^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] \angle ABO = \angle BAO = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle ABO + \angle BAO + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAB \\[3ex] \implies \\[3ex] p + p + 112 = 180 \\[3ex] 2p = 180 - 112 \\[3ex] 2p = 68 \\[3ex] p = \dfrac{68}{2} \\[5ex] p = 34 \\[3ex] \therefore \angle ABO = \angle BAO = 34^\circ $(23.) NSC In the diagram, O is the centre of the circle. Radii OH and OJ are drawn. A tangent os drawn from K to touch the circle at H.$\triangle$HGJ is drawn such that GJ || KH$\hat{O_1} = 2x$(23.1.1) Name, giving reasons, THREE angles, each equal to x (23.1.2) Prove that$\hat{H_2} = \hat{H_3} (23.1.1) \\[3ex] \underline{First\;\;angle\;\;equal\;\;to\;\;x} \\[3ex] \angle JOH = 2 * \angle JGH ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2x = 2 * \angle JGH \\[3ex] 2 * \angle JGH = 2x \\[3ex] \angle JGH = \dfrac{2x}{2} \\[5ex] \angle JGH = x \\[3ex] \underline{Second\;\;angle\;\;equal\;\;to\;\;x} \\[3ex] Parallel\;\;Lines:\;\;GJ\;\;and\;\;HK \\[3ex] \angle JGH = \angle GHK = x...alternate\;\;interior\;\; \angle s\;\;are\;\;equal \\[3ex] \underline{Third\;\;angle\;\;equal\;\;to\;\;x} \\[3ex] \angle GHK = \angle GJH = x ...\angle\;\;between\;\;tangent\;KH\;\;and\;\;chord\;GH = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (23.1.2) \\[3ex] \underline{\triangle HOJ} \\[3ex] \angle OHJ + \angle HOJ + \angle HJO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle HOJ \\[3ex] \hat{H_3} + 2x + \hat{H_1} = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle HOJ \\[3ex] \hat{H_3} = \hat{H_1} ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle HOJ \\[3ex] \hat{H_3} + 2x + \hat{H_3} = 180 \\[3ex] 2\hat{H_3} = 180 - 2x \\[3ex] \hat{H_3} = \dfrac{180 - 2x}{2} \\[5ex] \hat{H_3} = \dfrac{2(90 - x)}{2} \\[5ex] \hat{H_3} = 90 - x ...eqn.(1) \\[3ex] \underline{\triangle HGJ} \\[3ex] \angle GHJ + \angle JGH + \angle GJH = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle HGJ \\[3ex] (\hat{H_2} + \hat{H_3}) + x + x = 180 \\[3ex] \hat{H_2} + \hat{H_3} + 2x = 180 \\[3ex] \hat{H_2} = 180 - 2x - \hat{H_3} \\[3ex] From\;\;eqn.(1),\;\;substitute\;\;(90 - x)\;\;for\;\;\hat{H_3} \\[3ex] \hat{H_2} = 180 - 2x - (90 - x) \\[3ex] \hat{H_2} = 180 - 2x - 90 + x \\[3ex] \hat{H_2} = 90 - x...eqn.(2) \\[3ex] Comparing\;\;eqn.(1)\;\;and\;\;eqn.(2)\implies \\[3ex] \hat{H_2} = \hat{H_3} $(24.) CSEC In the diagram above, not drawn to scale,$O$is the centre of the circle.$\angle AOB = 130^\circ$,$\angle DAE = 30^\circ$, and$AEC$and$BED$are chords of the circle. (a) Calculate the size of EACH of the following angles, giving reasons for EACH step of your answers.$ (i)\:\: \angle ACB \\[3ex] (ii)\:\: \angle CBD \\[3ex] (iii)\:\: \angle AED \\[3ex]  (i) \\[3ex] \angle BOA = 2 * \angle ACB \\[3ex] ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 130 = 2 * \angle ACB \\[3ex] \angle ACB = \dfrac{130}{2} \\[5ex] \angle ACB = 65^\circ \\[3ex] (ii) \\[3ex] \angle CBD = \angle CAD = 30^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] (iii) \\[3ex] \angle ADB = \angle ACB = 65^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \underline{\triangle AED} \\[3ex] \angle DAE + \angle AED + \angle EDA = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle AED \\[3ex] \angle EDA = \angle ADB = 65 ...diagram \\[3ex] 30 + \angle AED + 65 = 180 \\[3ex] \angle AED + 95 = 180 \\[3ex] \angle AED = 180 - 95 \\[3ex] \angle AED = 85^\circ $(25.) WASSCE In the diagram,$\overline{RS}$and$\overline{RT}$are tangent to the circle with centre O$\angle TUS = 68^\circ$,$\angle SRT = x$and$\angle UTO = y$Find the value of$x \angle OTR = 90^\circ ...radius\;OT \perp tangent\;RT \;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle OSR = 90^\circ ...radius\;OS \perp tangent\;RS \;\;at\;\;point\;\;of\;\;contact\;S \\[3ex] \angle TOS = 2 * 68 ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] \angle TOS = 136^\circ \\[3ex] \underline{Quadrilateral\;\;TROS} \\[3ex] \angle T + \angle R + \angle S + \angle O = 360^\circ ...sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;Quadrilateral \\[3ex] 90 + x + 90 + 136 = 360 \\[3ex] x + 316 = 360 \\[3ex] x = 360 - 316 \\[3ex] x = 44^\circ $(26.) ACT In the figure below,$\overline{AC}$is a diameter of the circle, B is a point on the circle, and$\overline{AB} \cong \overline{BC}$What is the degree measure of$\angle ABC$?$ A.\;\; 45^\circ \\[3ex] B.\;\; 60^\circ \\[3ex] C.\;\; 75^\circ \\[3ex] D.\;\; 90^\circ \\[3ex] E.\;\;Cannot\;\;be\;\;determined\;\;from\;\;the\;\;given\;\;information \\[3ex]  \angle ABC = 90^\circ...\angle \;\;in\;\;a\;\;semicircle $(27.) CSEC In the diagram below, A, B, C and D are points on the circumference of a circle, with centre O AOC and BOD are diameters of the circle. AB and DC are parallel. (i.) State the reason why angle ABC is$90^\circ$(ii.) Determine the value of EACH of the following angles. Show detailed working where necessary and give a reason to support your answer. (a.) Angle BAC (b.) Angle q (iii.) Calculate the value of angle$r (i.) \\[3ex] \angle ABC = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] (ii.) \\[3ex] \underline{\triangle ABC} \\[3ex] 2p + 90 + 3p = 180 \\[3ex] 5p + 90 = 180 \\[3ex] 5p = 180 - 90 \\[3ex] 5p = 90 \\[3ex] p = \dfrac{90}{5} \\[5ex] p = 18 \\[3ex] (a.) \\[3ex] \angle BAC = 2p \\[3ex] \angle BAC = 2(18) \\[3ex] \angle BAC = 36^\circ \\[3ex] (b.) \\[3ex] q = 2p...\angle s\;\;in\;\;the\;\;same\;\;segment \\[3ex] q = 36^\circ \\[3ex] (iii.) \\[3ex] \underline{\triangle DOC} \\[3ex] r = q ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DOC \\[3ex] r = 36^\circ $(28.) ACT In the figure shown below, C, M, and N lie on the circle whose center is O, and$\angle MON$is a right angle. What is the sum of the measures of$\angle CMO$and$\angle CNO$?$ F.\;\; 90^\circ \\[3ex] G.\;\; 67.5^\circ \\[3ex] H.\;\; 60^\circ \\[3ex] J.\;\; 45^\circ \\[3ex] K.\;\; 22.5^\circ \\[3ex]  Center:\;\;O \\[3ex] \angle \;\;at\;\;center = 90^\circ \\[3ex] Let: \\[3ex] \angle MCN = \psi \\[3ex] \angle CMN = \theta \\[3ex] \angle CNM = \phi \\[3ex] \angle OMN = x \\[3ex] \angle ONM = y \\[3ex] 90 = 2 * \psi ...\angle \;\;at\;\;center = 2 * \angle\;\;at\;\;circumference \\[3ex] 2\psi = 90 \\[3ex] \psi = \dfrac{90}{2} \\[5ex] \psi = 45^\circ \\[3ex] Construction:\;\;Draw\;\;a\;\;chord\;\;to\;\;join\;\;M\;\;to\;\;N \\[3ex] \underline{\triangle OMN} \\[3ex] x = y ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OMN \\[3ex] Because\;\;of\;\;same\;\;radii...OM = ON \\[3ex] \implies \\[3ex] 90 + x + y = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OMN \\[3ex] 90 + x + x = 180 \\[3ex] 90 + 2x = 180 \\[3ex] 2x = 180 - 90 \\[3ex] 2x = 90 \\[3ex] x = \dfrac{90}{2} \\[5ex] x = 45 \\[3ex] x = y = 45^\circ \\[3ex] \underline{\triangle CMN} \\[3ex] \psi + \theta + x + y + \phi = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CMN \\[3ex] 45 + \theta + 45 + 45 + \phi = 180 \\[3ex] 135 + \theta + \phi = 180 \\[3ex] \theta + \phi = 180 - 135 \\[3ex] \theta + \phi = 45^\circ $(29.) WASSCE In the diagram,$O$is the centre of the circle.$AK$is a straight line and$TK$is a tangent. If$\angle CTK = 30^\circ$, calculate$\angle TKC \angle TAC = 30 ...\angle \:\:between\:\:tangent:\:TK\;\;and\:\:chord:\;TC = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle ATC = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle TKC} \\[3ex] \angle TCK = 90 + 30 ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle TCK = 120 \\[3ex] \angle TKC + \angle TCK + \angle CTK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle TKC + 120 + 30 = 180 \\[3ex] \angle TKC + 150 = 180 \\[3ex] \angle TKC = 180 - 150 \\[3ex] \angle TKC = 30^\circ $(30.) JAMB In the figure above, PQRS is a circle with ST || RQ Find the value of x if PT = PS$ A.\;\; 70^\circ \\[3ex] B.\;\; 55^\circ \\[3ex] C.\;\; 40^\circ \\[3ex] D.\;\; 35^\circ \\[3ex]  \angle STP = x ...corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;PQRS} \\[3ex] \angle P + \angle R = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle P + 110 = 180 \\[3ex] \angle P = 180 - 110 \\[3ex] \angle P = 70^\circ \\[3ex] \underline{\triangle PST} \\[3ex] \angle PST = \angle STP = x...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PST \\[3ex] \angle PST + \angle STP + \angle SPT = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle PST \\[3ex] \angle SPT = \angle P = 70^\circ ...diagram \\[3ex] \implies \\[3ex] x + x + 70 = 180 \\[3ex] 2x + 70 = 110 \\[3ex] 2x = 180 - 70 \\[3ex] 2x = 110 \\[3ex] x = \dfrac{110}{2} \\[5ex] x = 55^\circ $(31.) WASSCE In the diagram is a circle MNPR with centre O The angle at O is$196^\circ$and$\angle NMO = 52^\circ$Find the value of m$ \angle MOP + 196 = 360^\circ ...sum\;\;of\;\;\angle s\;\;at\;\;a\;\;point \\[3ex] \angle MOP = 360 - 196 \\[3ex] \angle MOP = 164^\circ \\[3ex] \angle MOP = 2 * \angle MNP ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 164 = 2 * \angle MNP \\[3ex] 2 * \angle MNP = 164 \\[3ex] \angle MNP = \dfrac{164}{2} \\[5ex] \angle MNP = 82^\circ \\[3ex] \underline{\triangle MOP} \\[3ex] \angle OMP = \angle OPM = p...base\;\;angle s\;\;of\;\;isosceles\;\;\triangle MOP \\[3ex] p + 164 + p = 180^\circ ...sum\;\;of\;\;of\;\;\angle s\;\;of\;\;\triangle MOP \\[3ex] 2p + 164 = 180 \\[3ex] 2p = 180 - 164 \\[3ex] 2p = 16 \\[3ex] p = \dfrac{16}{2} \\[5ex] p = 8^\circ \\[3ex] \underline{\triangle MNP} \\[3ex] \angle PMN + \angle MNP + \angle NPM = 180^\circ ...sum\;\;of\;\;of\;\;\angle s\;\;of\;\;\triangle MNP \\[3ex] (52 + p) + 82 + (m + p) = 180 \\[3ex] 52 + p + 82 + m + p = 180 \\[3ex] 52 + 8 + 82 + m + 8 = 180 \\[3ex] m + 150 = 180 \\[3ex] m = 180 - 150 \\[3ex] m = 30^\circ $(32.) GCSE P, Q, R and S are points on a circle. PXR and QXS are straight lines. PX = SX Prove that QS is not a diameter of the circle. If QS is the diameter of the circle, then$\angle QPS = 90^\circ...\angle\;\;in\;\;a\;\;semicircle$If QS is not the diameter of the circle, then$\angle QPS \ne 90^\circ$So, let us calculate$\angle QPS \angle QPR = 27^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle XPS} \\[3ex] \angle XPS = \angle XSP = p...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PXS \\[3ex] p + p + 50 = 180...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PXS \\[3ex] 2p + 50 = 180 \\[3ex] 2p = 180 - 50 \\[3ex] 2p = 130 \\[3ex] p = \dfrac{130}{2} \\[5ex] p = 65^\circ \\[3ex] \angle QPS = \angle QPR + \angle XPS...diagram \\[3ex] \angle QPS = 27 + 65 \\[3ex] \angle QPS = 92^\circ \\[3ex] 92^\circ \ne 90^\circ \\[3ex] $Therefore, QS is not a diameter of the circle. (33.) CSEC In the diagram below, not drawn to scale, O is the centre of the circle. The measure of angle LOM is$110^\circ$Calculate, giving reasons for your answers, the size of EACH of the following angles (i)$\angle MNL$(ii)$\angle LMO (i) \\[3ex] \angle MOL = 2 * \angle MNL ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] 110 = 2 * \angle MNL \\[3ex] 2 * \angle MNL = 110 \\[3ex] \angle MNL = \dfrac{110}{2} \\[5ex] \angle MNL = 55^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle MOL} \\[3ex] \angle LMO + \angle MLO + \angle MOL = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MOL \\[3ex] \angle LMO = \angle MLO = p ...base \angle s\;\;of\;\;isosceles\;\;\triangle MOL \\[3ex] \implies \\[3ex] p + p + 110 = 180 \\[3ex] 2p + 110 = 180 \\[3ex] 2p = 180 - 110 \\[3ex] 2p = 70 \\[3ex] p = \dfrac{70}{2} \\[5ex] p = 35 \\[3ex] \therefore \angle LMO = 35^\circ $(34.) GCSE A, B and C are points on the circumference of a circle with centre O$A\hat{C}B = 74^\circ$Calculate the value of x You must state any angle property of a circle that you use. You must show all your working.$ \angle AOB = 2 * \angle ACB ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] \angle AOB = 2 * 74 \\[3ex] \angle AOB = 148^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] \angle OBA = \angle OAB = x ...base \angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle OBA + \angle OAB + \angle AOB = ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] x + x + 148 = 180 \\[3ex] 2x + 148 = 180 \\[3ex] 2x = 180 - 148 \\[3ex] 2x = 32 \\[3ex] x = \dfrac{32}{2} \\[5ex] x = 16^\circ $(35.) WASSCE Calculate (i.)$\angle ODB$(ii.)$\angle BOD$Construction: Draw the chord from A to D$ (i) \\[3ex] \underline{\triangle AOD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] Let: \\[3ex] \angle OAD = ODA = p... base\;\; \angle s\;\;of\;\;isosceles\;\;\triangle AOD \\[3ex] \implies \\[3ex] p + p + 130 = 180 \\[3ex] 2p + 130 = 180 \\[3ex] 2p = 180 - 130 \\[3ex] 2p = 50 \\[3ex] p = \dfrac{50}{2} \\[5ex] p = 25 \\[3ex] \underline{\triangle ABD} \\[3ex] Let\;\; \angle ODB = k \\[3ex] \angle BAD + \angle ADB + \angle ABD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \angle BAD = 26 + p = 26 + 25 = 51^\circ ...diagram \\[3ex] \angle ADB = k + 25 ...diagram \\[3ex] \angle AOD = 2 * \angle ABD ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] 130 = 2 * \angle ABD \\[3ex] 2 * \angle ABD = 130 \\[3ex] \angle ABD = \dfrac{130}{2} \\[5ex] \angle ABD = 65^\circ \\[3ex] \implies \\[3ex] 51 + (k + 25) + 65 = 180 \\[3ex] 51 + k + 25 + 65 = 180 \\[3ex] 141 + k = 180 \\[3ex] k = 180 - 141 \\[3ex] k = 39 \\[3ex] \therefore \angle ODB = 39^\circ \\[3ex] (ii) \\[3ex] $Construction: Join O to B to complete the triangle BOD$ \underline{\triangle BOD} \\[3ex] \angle ODB = \angle OBD = k = 39 ... base \angle s\;\;of\;\;isosceles\;\;\triangle BOD \\[3ex] \angle ODB + \angle OBD + \angle BOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOD \\[3ex] 39 + 39 + \angle BOD = 180 \\[3ex] 78 + \angle BOD = 180 \\[3ex] \angle BOD = 180 - 78 \\[3ex] \angle BOD = 102^\circ $(36.) GCSE A, B and C are points on the circumference of a circle, centre O (a.) Write down the mathematical name for (i.) the straight line AC, (ii.) the straight line AB (b.) Give a geometrical reason why angle ABC = 90°$ (a.) \\[3ex] (i.) \\[3ex] Straight\;\;line\;AC \;\;is\;\;the\;\;diameter \\[3ex] (ii.) \\[3ex] Straight\;\;line\;AB \;\;is\;\;a\;\;chord \\[3ex] (b.) \\[3ex] \angle ABC = 90^\circ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle $(37.) GCSE A, B and C are points on the circumference of a circle, centre O DAE is the tangent to the circle at A Angle BAE =$56^\circ$Angle CBO =$35^\circ$Work out the size of angle CAO You must show all your working.$ \angle ACB = 56^\circ...\angle \;\;between\;\;tangent\;DAE\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle AOB = 2 * \angle ACB ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] \angle AOB = 2 * 56 \\[3ex] \angle AOB = 112^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] \angle BAO = \angle ABO = k ... base \angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle BAO + \angle ABO + \angle AOB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] k + k + 112 = 180 \\[3ex] 2k + 112 = 180 \\[3ex] 2k = 180 - 112 \\[3ex] 2k = 68 \\[3ex] k = \dfrac{68}{2} \\[5ex] k = 34 \\[3ex] \therefore \angle BAO = \angle ABO = 34^\circ \\[3ex] \underline{\triangle ACB} \\[3ex] \angle ABC = \angle ABO + \angle CBO ...diagram \\[3ex] \angle ABC = 34 + 35 \\[3ex] \angle ABC = 69^\circ \\[3ex] \angle BAC = \angle BAO + \angle CAO ...diagram \\[3ex] \angle BAC = 34 + \angle CAO \\[3ex] \angle ACB + \angle ABC + \angle BAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACB \\[3ex] 56 + 69 + (34 + \angle CAO) = 180 \\[3ex] 125 + 34 + \angle CAO = 180 \\[3ex] 159 + \angle CAO = 180 \\[3ex] \angle CAO = 180 - 159 \\[3ex] \angle CAO = 21^\circ $(38.) CSEC The diagram below, not drawn to scale, shows two circles. C is the centre of the smaller circle, GH is a common chord and DEF is a triangle. Angle GCH = 88° and angle GHE = 126° Calculate, giving reasons for your answer, the measure of angle (i) GFH (ii) GDE (iii) DEF$ (i) \\[3ex] \underline{Smaller\;\;Circle:\;\;Centre:\;C} \\[3ex] \angle GCH = 2 * \angle GFH ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 88 = 2 * \angle GFH \\[3ex] 2 * \angle GFH = 88 \\[3ex] \angle GFH = \dfrac{88}{2} \\[5ex] \angle GFH = 44^\circ \\[3ex] (ii) \\[3ex] \underline{Bigger\;\;Circle} \\[3ex] Cyclic\;\;Quadrilateral\;DGHE \\[3ex] \angle D + \angle H = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle D = \angle GDE \\[3ex] \angle H = \angle GHE = 126^\circ \\[3ex] \implies \\[3ex] \angle D + 126 = 180 \\[3ex] \angle D = 180 - 126 \\[3ex] \angle D = 54 \\[3ex] \angle GDE = 54^\circ \\[3ex] (iii) \\[3ex] \underline{Both\;\;Circles} \\[3ex] \triangle DFE \\[3ex] \angle DEF + \angle EDF + \angle DFE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DFE \\[3ex] \angle DEF = ? \\[3ex] \angle EDF = \angle GDE = 54^\circ \\[3ex] \angle DFE = \angle GFH = 44^\circ \\[3ex] \implies \\[3ex] \angle DEF + 54 + 44 = 180 \\[3ex] \angle DEF + 98 = 180 \\[3ex] \angle DEF = 180 - 98 \\[3ex] \angle DEF = 82^\circ $(39.) curriculum.gov.mt The diagram shows a circle with centre C. Two tangents AT and BT meet outside the circle at point T. (a) Triangle CBT is right-angled. Why? (b) Show that triangles CAR and CBT are congruent. (c) Explain why ACBT is a cyclic quadrilateral.$ (a) \\[3ex] \triangle CBT\;\;is\;\;right-angled\;\;because \\[3ex] \angle CBT = 90^\circ ... radius\;CB \perp tangent\;BT \;\;at\;\;point\;\;of\;\;contact\; B \\[5ex] (b) \\[3ex] At\;\;least\;\;three\;\;ways\;\;to\;\;show\;\;Congruency \\[3ex] Side:\;\; CA = CB ...radius \\[3ex] Angle:\;\; \angle CAT = \angle CBT = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AT = BT ...two\;\;tangents:\;\;AT\;\;and\;\;BT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \triangle CAT \cong \triangle CBT...Side-Angle-Side \\[3ex] OR \\[3ex] Angle:\;\; \angle CAT = \angle CBT = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AT = BT ...two\;\;tangents:\;\;AT\;\;and\;\;BT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] Angle:\;\; \angle ATC = \angle BTC ... \\[3ex] $For two tangents: AT and BT drawn from the same external point: T, the line joining the external point: T, and the centre of the circle: C; bisects the angle between the tangents.$ \implies \\[3ex] \triangle CAT \cong \triangle CBT...Angle-Side-Angle \\[3ex] OR \\[3ex] Side:\;\; CA = CB ...radius \\[3ex] Side:\;\; AT = BT ...two\;\;tangents:\;\;AT\;\;and\;\;BT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] Side:\;\; CT = CT ... common\;\;side\;\;of\;\;the\;\;two\;\;\triangle s \\[3ex] \implies \\[3ex] \triangle CAT \cong \triangle CBT...Side-Side-Side \\[5ex] (c) \\[3ex] Because\;\;the\;\;polygon\;ACBT\;\;is\;\;a\;\;quadrilateral \\[3ex] where \\[3ex] \angle CAT = \angle CBT = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] and \\[3ex] \angle CAT \;\;and\;\; \angle CBT \;\;are\;\;opposite\;\;angles \\[3ex] \implies \\[3ex] Quadrilateral\;\;ACBT\;\;is\;\;a\;\;cyclic\;\;quadrilateral \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary\;\;(90^\circ + 90^\circ = 180^\circ) $(40.) WASSCE In the diagram, O is the centre of the circle.$\angle ACB = 39^\circ$and$\angle CBE = 62^\circ$Find: (i.) the interior angle AOC (ii.) angle BAC$ \angle ABC + 62 = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ABC = 180 - 62 \\[3ex] \angle ABC = 118^\circ \\[3ex] Reflex\;\; \angle AOC = 2 * \angle ABC \\[3ex] ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] Reflex\;\; \angle AOC = 2 * 118 \\[3ex] Reflex\;\; \angle AOC = 236^\circ \\[3ex] (i.) \\[3ex] interior\;\;\angle AOC + Reflex\;\;\angle AOC = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] interior\;\;\angle AOC + 236 = 360 \\[3ex] interior\;\;\angle AOC = 360 - 236 \\[3ex] interior\;\;\angle AOC = 124^\circ \\[3ex] (ii.) \\[3ex] \underline{\triangle ABC} \\[3ex] \angle BAC + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle BAC + 118 + 39 = 180 \\[3ex] \angle BAC + 157 = 180 \\[3ex] \angle BAC = 180 - 157 \\[3ex] \angle BAC = 23^\circ $(41.) GCSE The diagram shows a circle, centre O. Points A, B, C and D lie on the circumference of the circle. EDF is a tangent to the circle. Angle ABC = 82° and angle ODC = 57° (a.) Work out the value of x. (b.) Work out the value of y.$ (a.) \\[3ex] \underline{\triangle DOC} \\[3ex] \angle ODC = \angle OCD = 57^\circ ... base\;\; \angle s\;\;of\;\;isosceles\;\;\triangle DOC \\[3ex] \angle ODC + \angle OCD + \angle DOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DOC \\[3ex] 57 + 57 + x = 180 \\[3ex] 114 + x = 180 \\[3ex] x = 180 - 114 \\[3ex] x = 66^\circ \\[3ex] (b.) \\[3ex] Let\;\;\angle ADO = p \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle B = 82^\circ \\[3ex] \angle D = \angle ADO + \angle ODC ...diagram \\[3ex] \angle D = p + 57 \\[3ex] \angle D + \angle B = 180 ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \implies \\[3ex] (p + 57) + 82 = 180 \\[3ex] p + 57 + 82 = 180 \\[3ex] p + 139 = 180 \\[3ex] p = 180 - 139 \\[3ex] p = 41^\circ \\[3ex] \angle ODF = 90^\circ ...radius\;OD \perp tangent\;EDF\;\;at\;\;point\;\;of\;\;contact\;D \\[3ex] y + \angle ADO + \angle ODF = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] y + p + 90 = 180 \\[3ex] y + 41 + 90 = 180 \\[3ex] y + 131 = 180 \\[3ex] y = 180 - 131 \\[3ex] y = 49^\circ $(42.) CSEC The diagram, below, not drawn to scale, shows a circle, centre, O SGH is a tangent to the circle,$\angle FOG = 118^\circ$and$\angle DGS = 65^\circ$Calculate, giving reasons for EACH step of your answer, the measure of: (i)$\angle OGF$(ii)$\angle DEF (i) \\[3ex] \underline{\triangle GOF} \\[3ex] \angle OGF = \angle OFG = p...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle GOF \\[3ex] \angle OGF + \angle OFG + \angle GOF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle GOF \\[3ex] p + p + 118 = 180 \\[3ex] 2p = 180 - 118 \\[3ex] 2p = 62 \\[3ex] p = \dfrac{62}{2} \\[5ex] p = 31 \\[3ex] \therefore \angle OGF = 31^\circ \\[3ex] (ii) \\[3ex] \angle OGH = 90^\circ \\[3ex] ...radius\;OG\;\;\perp\;\;tangent\;SGH\;\;at\;\;point\;\;of\;\;contact\;G \\[3ex] \angle SGD + \angle DGO + \angle OGH = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 65 + \angle DGO + 90 = 180 \\[3ex] \angle DGO + 155 = 180 \\[3ex] \angle DGO = 180 - 155 \\[3ex] \angle DGO = 25^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;GDEF} \\[3ex] \angle G + \angle E = 180 \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle G = \angle DGO + \angle OGF \\[3ex] \angle G = 25 + 31 \\[3ex] \angle G = 56^\circ \\[3ex] \angle E = \angle DEF \\[3ex] \implies \\[3ex] 56 + \angle E = 180 \\[3ex] \angle E = 180 - 56 \\[3ex] \angle E = 124 \\[3ex] \therefore \angle DEF = 124^\circ $(43.) Did you notice anything in this question? If yes, what did you notice? If no, add the two digits of this question number. Check out the question number that has the sum. CSEC The diagram below, not drawn to scale, shows a circle with centre$O$.$HJ$and$HG$are tangents to the circle and$\angle AEB = 48^\circ$Calculate, giving reasons for your answer, the size of EACH of the following angles:$ (i)\:\: \angle OAE \\[3ex] (ii)\:\: \angle AOB \\[3ex] (iii)\:\: \angle ACB \\[3ex] (iv)\:\: \angle ADB \\[3ex]  (i) \\[3ex] \angle OAE = 90^\circ ... radius\;OA \perp tangent\;AE\:\:at\:\:point\:\:of\:\:contact\;A \\[3ex] (ii) \\[3ex] \angle OBE = 90^\circ ... radius\;OB \perp tangent\:BE\:\:at\:\:point\:\:of\:\:contact\;B \\[3ex] \underline{Quadrilateral\:\:AOBE} \\[3ex] \angle A + \angle O + \angle B + \angle E = 360 ...sum\:\:of\:\:interior\:\:\angle s\:\:of\:\:a\:\:Quadrilateral \\[3ex] 90 + \angle O + 90 + 48 = 360 \\[3ex] 228 + \angle O = 360 \\[3ex] \angle O = 360 - 228 \\[3ex] \angle O = 132 \\[3ex] \angle O = \angle AOB = 132^\circ \\[3ex] (iii) \\[3ex] Obtuse\;\;\angle AOB = 2 * \angle ACB ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 132 = 2 * \angle ACB \\[3ex] 2 * \angle ACB = 132 \\[3ex] \angle ACB = \dfrac{132}{2} \\[5ex] \angle ACB = 66^\circ \\[3ex] (iv) \\[3ex] Reflex\;\;\angle AOB + Obtuse\;\;\angle AOB = 360 ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle AOB + 132 = 360 \\[3ex] Reflex\;\;\angle AOB = 360 - 132 \\[3ex] Reflex\;\;\angle AOB = 228 \\[3ex] Reflex\;\;\angle AOB = 2 * \angle ADB ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 228 = 2 * \angle ADB \\[3ex] 2 * \angle ADB = 228 \\[3ex] \angle ADB = \dfrac{228}{2} \\[5ex] \angle ADB = 114^\circ \\[3ex] OR \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:ACBD} \\[3ex] \angle C + \angle D = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral \\[3ex] 66 + \angle D = 180 \\[3ex] \angle D = 180 - 66 \\[3ex] \angle D = 114 \\[3ex] \angle D = \angle ADB = 114^\circ $(44.) GCSE A, B, C and D are points on the circumference of the circle, centre O DOB is a straight line and angle DAC = 58° Find angle CDB$ \angle DBC = 58^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle DCB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;angle \\[3ex] \underline{\triangle DCB} \\[3ex] \angle CDB + \angle DCB + \angle DBC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DCB \\[3ex] \angle CDB + 90 + 58 = 180 \\[3ex] \angle CDB + 148 = 180 \\[3ex] \angle CDB = 180 - 148 \\[3ex] \angle CDB = 32^\circ $(45.) GCSE TV is a tangent to the circle at P SR = RQ Angle QPV = 41° and angle SQP = 57° Show that SP is parallel to RQ You must give reasons to justify any angles that you calculate. We can review the properties of angles between parallel lines to determine that SP is parallel to RQ Of those properties, the easiest one to check for is the Alternate Interior Angles property...alternate interior angles between parallel lines are equal Let us demonstrate it with a diagram So, we need to show that$\angle RQS = \angle PSQ$...indicated by the angles marked in red Once, we can show that$\angle RQS = \angle PSQ$, then we have shown one of the properties of angles between parallel lines and hence show that the two lines: SP and RQ are parallel$ \angle PSQ = 41^\circ ...\angle \;\;between\;\;tangent\;TPV\;\;and\;\;chord\;PS = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \underline{\triangle PSQ} \\[3ex] \angle SPQ + \angle PSQ + \angle PQS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] \angle SPQ + 41 + 57 = 180 \\[3ex] \angle SPQ + 98 = 180 \\[3ex] \angle SPQ = 180 - 98 \\[3ex] \angle SPQ = 82^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;SRQP} \\[3ex] \angle P + \angle R = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle P = \angle SPQ = 82^\circ \\[3ex] \implies \\[3ex] 82 + \angle R = 180 \\[3ex] \angle R = 180 - 82 \\[3ex] \angle R = 98^\circ \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ = \angle RQS = p ...isosceles\;\;\triangle SRQ \\[3ex] \angle SRQ = \angle R = 98^\circ \\[3ex] \angle RSQ + \angle RQS + \angle SRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] p + p + 98 = 180 \\[3ex] 2p = 180 - 98 \\[3ex] 2p = 82 \\[3ex] p = 41 \\[3ex] \therefore \angle RQS = 41^\circ \\[3ex] Because\;\; \angle PSQ = \angle RQS = 41^\circ,\;\;SP\;\;||\;\;RQ $(46.) JAMB In the diagram abovem POQ is a diameter, O is the centre of the circle and TP s a tangent. Find the value of x$ A.\;\; 30^\circ \\[3ex] B.\;\; 40^\circ \\[3ex] C.\;\; 45^\circ \\[3ex] D.\;\; 50^\circ \\[3ex]  \angle OPT = 90^\circ \\[3ex] ...radius\;OP\perp tangent\;PT\;\;at\;\;point\;\;of\;\;contact\;P \\[3ex] \underline{\triangle OPT} \\[3ex] \angle POT + \angle OPT + \angle OTP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPT \\[3ex] \angle POT + 90 + 30 = 180 \\[3ex] \angle POT + 120 = 180 \\[3ex] \angle POT = 180 - 120 \\[3ex] \angle POT = 60^\circ \\[3ex] \underline{\triangle OPR} \\[3ex] \angle OPR = \angle ORP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OPR \\[3ex] \angle OPR + \angle ORP + \angle POR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPR \\[3ex] \angle POR = \angle POT = 60^\circ \\[3ex] \implies \\[3ex] k + k + 60 = 180 \\[3ex] 2k = 180 - 60 \\[3ex] 2k = 120 \\[3ex] k = \dfrac{120}{2} \\[5ex] k = 60 \\[3ex] \therefore \angle OPR = \angle ORP = 60^\circ \\[3ex] \triangle OPR \;\;is\;\;an\;\;equilateral\;\;triangle \\[3ex] \angle QRP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \underline{\triangle PQR} \\[3ex] \angle PQR + \angle QRP + \angle QPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] \angle QPR = \angle OPR = 60^\circ \\[3ex] \implies \\[3ex] x + 90 + 60 = 180 \\[3ex] x + 150 = 180 \\[3ex] x = 180 - 150 \\[3ex] x = 30^\circ $(47.) NSC In the diagram, KLMN is a cyclic quadrilateral with$K\hat{L}M = 87^\circ$Diagonals LN and MK are drawn. P is a point on the circle and MP is produced to T, a point outside the circle. Chord LP is drawn.$L\hat{M}K = y$and$\hat{N_1} = 2y$(47.1) Name, giving a reason, another angle equal to y (47.2) Calculate, giving reasons, the size of: (a) y (b)$T\hat{P}L (47.1) \\[3ex] \angle LNK = \hat{N_2} = y...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (47.2) \\[3ex] (a) \\[3ex] \angle LKM = \hat{K_1} = 2y ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle LMK} \\[3ex] \angle KLM + \angle LKM + \angle LMK = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle LMK \\[3ex] 87 + 2y + y = 180 \\[3ex] 3y = 180 - 87 \\[3ex] 3y = 93 \\[3ex] y = \dfrac{93}{3} \\[5ex] y = 31^\circ \\[3ex] (b) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PMNL} \\[3ex] T\hat{P}L = \hat{K_1} = 2y ...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quad = interior\;\;opposite\;\;\angle \\[3ex] T\hat{P}L = 2(31) \\[3ex] T\hat{P}L = 62^\circ $(48.) curriculum.gov.mt Find the angles marked r, s and t in the diagrams below. Give reasons for your answers. Note: Diagrams are not drawn to scale.$ (a) \\[3ex] r = 30^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (b) \\[3ex] s + 68 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] s = 180 - 68 \\[3ex] s = 112^\circ \\[3ex] (c) \\[3ex] The\;\;third\;\;angle = 90^\circ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] t + 63 + 90 = 180 ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle \\[3ex] t + 153 = 180 \\[3ex] t = 180 - 153 \\[3ex] t = 27^\circ $(49.) GCSE (a.) A, B, C and D are points on a circle. The line PQ is a tangent to the circle at B$P\hat{B}A = 51^\circ$,$B\hat{C}D = 118^\circ$and$A\hat{B}D = w^\circ$Find the value of w You must show all your working. (b.) E, F and G are points on a circle with centre H$E\hat{F}H = x$and$G\hat{F}H = y$Complete the proof of the following statement: Tha angle at the centre is twice the angle at the circumference. Proof:$E\hat{F}G = x + yF\hat{E}H = x$(triangle FEH is isosceles)$F\hat{H}E = ................................. (a.) \\[3ex] 51 + w = 118 ...\angle \:\:between\:\:tangent:\:PQ\;\;and\:\:chord:\;BD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] w = 118 - 51 \\[3ex] w = 67^\circ \\[3ex] (b.) \\[3ex] For\;\;this\;\;question: \\[3ex] \underline{Given}: \\[3ex] E\hat{F}G = x + y ...\angle \;\;at\;\;circumference \\[3ex] Acute\;\; \angle EHG = \angle \;\;at\;\;centre \\[3ex] \underline{To\;\;Prove}: \\[3ex] \angle EHG = 2(x + y) \\[3ex] \underline{\triangle FEH} \\[3ex] E\hat{F}H = x...as\;\;shown\;\;in\;\;the\;\;diagram \\[3ex] F\hat{E}H = E\hat{F}H = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] F\hat{H}E + F\hat{E}H + H\hat{F}E = 180...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FEH \\[3ex] F\hat{H}E + x + x = 180 \\[3ex] F\hat{H}E + 2x = 180 \\[3ex] F\hat{H}E = 180 - 2x ...eqn.(1) \\[3ex] \underline{\triangle FEG} \\[3ex] H\hat{F}G = y...as\;\;shown\;\;in\;\;the\;\;diagram \\[3ex] F\hat{G}H = H\hat{F}G = y ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] F\hat{H}G + F\hat{G}H + H\hat{F}G = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FEH \\[3ex] F\hat{H}G + y + y = 180 \\[3ex] F\hat{H}G + 2y = 180 \\[3ex] F\hat{H}G = 180 - 2y ...eqn.(2) \\[3ex] Reflex\;\; \angle EHG = F\hat{H}E + F\hat{H}G \\[3ex] Reflex\;\; \angle EHG = (180 - 2x) + (180 - 2y) \\[3ex] Reflex\;\; \angle EHG = 180 - 2x + 180 - 2y \\[3ex] Reflex\;\; \angle EHG = 360 - 2x - 2y \\[3ex] Reflex\;\; \angle EHG + Acute\;\; \angle EHG = 360^\circ...sum\;\;of\;\;\angle s\;\;at\;\;a\;\;point \\[3ex] (360 - 2x - 2y) + Acute\;\; \angle EHG = 360 \\[3ex] Acute\;\;\angle EHG = 360 - (360 - 2x - 2y) \\[3ex] Acute\;\;\angle EHG = 360 - 360 + 2x + 2y \\[3ex] Acute\;\;\angle EHG = 2x + 2y \\[3ex] Acute\;\;\angle EHG = 2(x + y) \\[3ex] \implies \\[3ex] $The angle at the centre is twice the angle at the circumference. (50.) curriculum.gov.mt a) Prove that in a circle, the angles in the same segment are equal. Diagram NOT to scale i) p = ........°; Reason: ................................... ii) Calculate the value of angle q, giving reasons for every step. Construction: Draw the radii: from point P on the circumference to point O on the center; and from point Q on the circumference to point O on the center.$ a) \\[3ex] \angle POQ = 2 * \angle PAQ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Also: \\[3ex] \angle POQ = 2 * \angle PBQ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 2 * \angle PAQ = 2 * \angle PBQ \\[3ex] Divide\;\;both\;\;sides\;\;by\;\;2 \\[3ex] \angle PAQ = \angle PBQ \\[5ex] b) \\[3ex] i) \\[3ex] p = 65^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] ii) \\[3ex] 80^\circ = \angle DAB ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;ABCD = interior\;\;opposite\;\;\angle \\[3ex] \angle DAB = 80^\circ \\[3ex] q + \angle DAB + 45 = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle DAB \\[3ex] q + 80 + 45 = 180 \\[3ex] q + 125 = 180 \\[3ex] q = 180 - 125 \\[3ex] q = 55^\circ $(51.) WASSCE In the diagram, O is the centre of the circle |PQ| = |QR| and ∠PSR = 56° Find ∠QRS Construction: Join the line from Point P to Point R$ \angle PRS = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \underline{Cyclic\;\;Quadrilateral\;QPSR} \\[3ex] \angle Q + \angle S = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle Q + 56 = 180 \\[3ex] \angle Q = 180 - 56 \\[3ex] \angle Q = 124^\circ \\[3ex] \underline{\triangle RQP} \\[3ex] \angle QRP = \angle QPR = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RQP \\[3ex] \angle Q = \angle RQP = 124^\circ ...diagram \\[3ex ] \angle QRP + \angle QPR + \angle RQP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RQP \\[3ex] k + k + 124 = 180 \\[3ex] 2k = 180 - 124 \\[3ex] 2k = 56 \\[3ex] k = \dfrac{56}{2} \\[5ex] k = 28 \\[3ex] \therefore \angle QRP = \angle QPR = 28^\circ \\[3ex] \angle QRS = \angle QRP + \angle PRS ...diagram \\[3ex] \angle QRS = 28 + 90 \\[3ex] \angle QRS = 118^\circ $(52.) NZQA Below is part of a spider web with a hole in it. Points P, Q, and R all lie on the circumference of a circle, with centre C Angle PRC = 25° Calculate the size, a, of angle PQR Justify your answer.$ \underline{\triangle PCR} \\[3ex] \angle RPC = \angle PRC = 25^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PCR \\[3ex] \angle PCR + \angle RPC + \angle PRC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PCR \\[3ex] \angle PCR + 25 + 25 = 180 \\[3ex] \angle PCR + 50 = 180 \\[3ex] \angle PCR = 180 - 50 \\[3ex] \angle PCR = 130^\circ \\[3ex] \angle PCR = 2 * \angle PQR ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] 130 = 2 * \angle PQR \\[3ex] 2 * \angle PQR = 130 \\[3ex] \angle PQR = \dfrac{130}{2} \\[5ex] \angle PQR = 65 \\[3ex] \angle PQR = a = 65^\circ $(53.) CSEC The diagram below, not drawn to scale, shows a circle, centre O RQ is a diameter and PM and PN are tangents to the circle Angle MPN = 54° and angle RQM = 20° Calculate, giving reasons for your answer, the measure of: (i) ∠MRQ (ii) ∠PMR (iii) ∠PMN$ \angle RMQ = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle RMQ} \\[3ex] \angle MRQ + \angle RMQ + \angle RQM = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RMQ \\[3ex] \angle MRQ + 90 + 20 = 180 \\[3ex] \angle MRQ + 110 = 180 \\[3ex] \angle MRQ = 180 - 110 \\[3ex] \angle MRQ = 70^\circ \\[3ex] (ii) \\[3ex] \angle PMR = 20^\circ...\angle \;\;between\;\;tangent\;MP\;\;and\;\;chord\;MR = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (iii) \\[3ex] \underline{\triangle PMN} \\[3ex] |MP| = |NP|...two\;\;tangents:\;\;MP\;\;and\;\;NP\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \;\; \angle PMN = \angle PNM \\[3ex] \implies \;\; \triangle PMN \;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] Let\;\; \angle PMN = \angle PNM = k \\[3ex] \angle PMN + \angle PNM + \angle MPN = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PMN \\[3ex] k + k + 54 = 180 \\[3ex] 2k = 180 - 54 \\[3ex] 2k = 126 \\[3ex] k = \dfrac{126}{2} \\[5ex] k = 63 \\[3ex] \therefore \angle PMN = \angle PNM = 63^\circ $(54.) ACT A circle with center C is shown below. Points W, X, Y, and Z lie on the circle. The measure of$\angle WCY$is$100^\circ$, the measure of$\angle XCZ$is$80^\circ$, and the measure of$\angle WCZ$is$150^\circ$What is the measure of$\angle XCY$?$ A.\;\; 20^\circ \\[3ex] B.\;\; 25^\circ \\[3ex] C.\;\; 30^\circ \\[3ex] D.\;\; 40^\circ \\[3ex] E.\;\; 50^\circ \\[3ex] $The ACT is a timed test. It is important to do this question within a minute. Remember what I wrote earlier So, the faster approach for me is to use alphabets rather than angles This is what I mean$ Let: \\[3ex] \angle WCX = m \\[3ex] \angle XCY = n \\[3ex] \angle YCZ = p \\[3ex] Based\;\;on\;\;the\;\;question\;\;and\;\;modified\;\;diagram: \\[3ex] m + n = 100 ...eqn.(1) \\[3ex] n + p = 80...eqn.(2) \\[3ex] m + n + p = 150 ...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;into\;\;eqn.(3) \implies \\[3ex] 100 + p = 150 \\[3ex] p = 150 - 100 \\[3ex] p = 50 ...eqn.(4) \\[3ex] Substitute\;\;eqn.(4)\;\;into\;\;eqn.(2) \implies \\[3ex] n + 50 = 80 \\[3ex] n = 80 - 50 \\[3ex] n = 30 \\[3ex] \therefore \angle XCY = 30^\circ $(55.) WASSCE In the diagram, QOS is a diameter,$\angle RQS = x^\circ$and$\angle QST = (3x + 15)^\circ$Find: (i.) the value of$x$(ii.)$\angle RSQ (i.) \\[3ex] \angle QRS = 90^\circ...\angle s\;\;in\;\;a\;\;semicircle \\[3ex] \angle QST = \angle RQS + \angle QRS \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s\\[3ex] \implies \\[3ex] (3x + 15) = x + 90 \\[3ex] 3x + 15 = x + 90 \\[3ex] 3x - x = 90 - 15 \\[3ex] 2x = 75 \\[3ex] x = \dfrac{75}{2} \\[5ex] x = 37.5^\circ \\[3ex] (ii.) \\[3ex] \angle RSQ + \angle QST = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RSQ + (3x + 15) = 180 \\[3ex] \angle RSQ + [3(37.5) + 15] = 180 \\[3ex] \angle RSQ + (112.5 + 15) = 180 \\[3ex] \angle RSQ + 127.5 = 180 \\[3ex] \angle RSQ = 180 - 127.5 \\[3ex] \angle RSQ = 52.5^\circ $(56.) LSSCE MA 11 Find the value of the pro-numeral b in degrees for the diagram below. O is the centre of the circle$ \angle O = 2 * b ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 72 = 2 * b \\[3ex] 2b = 72 \\[3ex] b = \dfrac{72}{2} \\[5ex] b = 36^\circ $(57.) WASSCE In the diagram, TS is a tangent to the circle at S |PR| = |RS| and ∠PQR = 117° Calculate ∠PST$ A.\;\; 54^\circ \\[3ex] B.\;\; 44^\circ \\[3ex] C.\;\; 34^\circ \\[3ex] D.\;\; 27^\circ \\[3ex]  \angle S + \angle Q = 180^\circ ...opposite\;\;\angle s\;\;of\;\;Cyclic\;\;Quadrilateral\;PQRS\;\;are\;\;supplementary \\[3ex] \angle Q = \angle PQR = 117^\circ ...diagram \\[3ex] \angle S = \angle RSP ...diagram \\[3ex] \implies \\[3ex] \angle RSP + 117 = 180 \\[3ex] \angle RSP = 180 - 117 \\[3ex] \angle RSP = 63^\circ \\[3ex] \angle RPS = \angle RSP = 63^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PRS \\[3ex] \angle PRS + \angle RPS + \angle RSP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] \angle PRS + 63 + 63 = 180 \\[3ex] \angle PRS + 126 = 180 \\[3ex] \angle PRS = 180 - 126 \\[3ex] \angle PRS = 54^\circ \\[3ex] \angle PST = \angle PRS = 54^\circ ...\angle\;\;between\;\;tangent\;TS\;\;and\;\;chord\;SP = \angle\;\;in\;\;the\;\;alternate\;\;segment $(58.) CSEC The diagram below, not drawn to scale, shows a circle with centre O TAE is a tangent to the circle at point A and angle AOD = 72° Calculate, giving the reason for each step of your answer, the measure of:$ (i)\;\; \angle ADC = \;\;\;................ \\[3ex] (ii)\;\; \angle ACD = \;\;\;................ \\[3ex] (iii)\;\; \angle CAD = \;\;\;................ \\[3ex] (iv)\;\; \angle OEA = \;\;\;................ \\[3ex]  (i) \\[3ex] \angle ADC = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] (ii) \\[3ex] \angle AOD = 2 * \angle ACD ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 72 = 2 * \angle ACD \\[3ex] 2 * \angle ACD = 72 \\[3ex] \angle ACD = \dfrac{72}{2} \\[5ex] \angle ACD = 36^\circ \\[3ex] OR \\[3ex] \underline{\triangle AOD} \\[3ex] \angle OAD = \angle ODA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] p + p + 72 = 180 \\[3ex] 2p = 180 - 72 \\[3ex] 2p = 108 \\[3ex] p = \dfrac{108}{2} \\[5ex] p = 54 \\[3ex] \therefore \angle OAD = \angle ODA = 54^\circ \\[3ex] \underline{\triangle ACD} \\[3ex] \angle CAD = \angle OAD = 54^\circ \\[3ex] \angle CAD + \angle ACD + \angle ADC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] 54 + \angle ACD + 90 = 180 \\[3ex] \angle ACD + 144 = 180 \\[3ex] \angle ACD = 180 - 144 \\[3ex] \angle ACD = 36^\circ \\[3ex] (iii) \\[3ex] \angle CAD = 54^\circ \\[3ex] (iv) \\[3ex] \angle OAE = 90^\circ \\[3ex] ... radius \;OA \; \perp \; tangent\;TAE\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \underline{\triangle OEA} \\[3ex] \angle OEA + \angle AOE + \angle OAE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OEA \\[3ex] \angle OEA + 72 + 90 = 180 \\[3ex] \angle OEA + 162 = 180 \\[3ex] \angle OEA = 180 - 162 \\[3ex] \angle OEA = 18^\circ $(59.) NSC (59.1) Complete the following theorem: Angles subtended by a chord of a circle, ..., are equal. (59.2) In the diagram below, O is the centre of circle KLMNP. KOM is a diameter of the circle and chords LP and PM are drawn.$\hat{N} = 128^\circ$Determine, stating reasons, the size of EACH of the following angles: (57.2.1)$\hat{K_1}$(57.2.2)$\hat{L_2}$(57.2.3)$\hat{P_2}$is$\hat{P_3} = 29^\circ$(57.1) Angles subtended by a chord of a circle, in the same segment of the circle, are equal.$ (57.2.1) \\[3ex] Reflex\;\;\angle O = 2 * \angle PNM ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle O = 2 * 128 \\[3ex] Reflex\;\;\angle O = 256^\circ \\[3ex] Reflex\;\;\angle O + Obtuse\;\;\angle O = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] 256 + Obtuse\;\;\angle O = 360 \\[3ex] Obtuse\;\;\angle O = 360 - 256 \\[3ex] Obtuse\;\;\angle O = 104^\circ \\[3ex] \underline{\triangle POM} \\[3ex] \angle OPM = \angle OMP = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POM \\[3ex] \angle OPM + \angle OMP + \angle POM = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POM \\[3ex] \angle POM = \hat{O_2} = Obtuse\;\;\angle O = 104^\circ \\[3ex] \implies \\[3ex] x + x + 104 = 180 \\[3ex] 2x = 180 - 104 \\[3ex] 2x = 76 \\[3ex] x = \dfrac{76}{2} \\[5ex] x = 38 \\[3ex] \therefore \angle OPM = \angle OMP = 38^\circ \\[3ex] \angle KPM = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle KMP} \\[3ex] \angle PKM + \angle KPM + \angle KMP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle KMP \\[3ex] \angle KMP = \hat{M_2} = \angle OMP = 38^\circ \\[3ex] \implies \\[3ex] \angle PKM + 90 + 38 = 180 \\[3ex] \angle PKM + 128 = 180 \\[3ex] \angle PKM = 180 - 128 \\[3ex] \angle PKM = 52 \\[3ex] \angle PKM = \hat{K_1} = 52^\circ \\[3ex] (57.2.2) \\[3ex] \hat{L_2} = \hat{M_2} = 38^\circ ...\angle s\;\;in\;\;the\;\;segment\;\;are\;\;equal \\[3ex] OR \\[3ex] \hat{L_1} = \hat{K_1} = 52^\circ ...\angle s\;\;in\;\;the\;\;segment\;\;are\;\;equal \\[3ex] \angle KLM = \hat{L} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \hat{L_1} + \hat{L_2} = \hat{L} ...diagram \\[3ex] 52 + \hat{L_2} = 90 \\[3ex] \hat{L_2} = 90 - 52 \\[3ex] \hat{L_2} = 38^\circ \\[3ex] (57.2.3) \\[3ex] \hat{P_2} + \hat{P_3} = \angle OPM ...diagram \\[3ex] \hat{P_2} + 29 = 38 \\[3ex] \hat{P_2} = 38 - 29 \\[3ex] \hat{P_2} = 9^\circ $(60.) NZQA In this spider web, points G, H, T, and S all lie on the circumference of a Circle, with centre C The straight lines FSJ and FTK are both tangents to the circle at the points S and T Angle SCT = 100° Determine whether the line FSJ is parallel to the line GCT Justify your answer with clear geometrical reasoning$ If\;\;|FSJ|\;\;\parallel\;\;|GCT|,\;\;then \\[3ex] \angle CSJ = 100^\circ ...alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] Let\;\;us\;\;calculate\;\;\angle CSJ \\[3ex] If\;\;\angle CSJ = 100^\circ,\;\;then\;\; |FSJ|\;\;\parallel\;\;|GCT| \\[3ex] If\;\;\angle CSJ \ne 100^\circ,\;\;then\;\; |FSJ|\;\;\nparallel\;\;|GCT| \\[3ex] \angle CSJ = 90^\circ \\[3ex] ...radius\;CS \;\;\perp\;\;tangent\;FSJ\;\;at\;\;point\;\;of\;\;contact\;S \\[3ex] \angle CSJ \ne 100^\circ \\[3ex] \therefore |FSJ|\;\;\nparallel\;\;|GCT| $(61.) GCSE PQRS is a cyclic quadrilateral of a circle, centre O.$S\hat{P}Q = 54^\circ$The ratio of$O\hat{S}R$to$O\hat{Q}R$is 2:1 (i) Find$S\hat{O}Q$Give a reason for your answer. (ii) Find$S\hat{R}Q$(iii) Calculate the size of$O\hat{S}R (i) \\[3ex] Obtuse\;\;S\hat{O}Q = 2 * \angle S\hat{P}Q ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;S\hat{O}Q = 2 * 54 \\[3ex] Obtuse\;\;S\hat{O}Q = 108^\circ \\[3ex] (ii) \\[3ex] Obtuse\;\;S\hat{O}Q + Reflex\;\;S\hat{O}Q = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] 108 + Reflec\;\;S\hat{O}Q = 360 \\[3ex] Reflex\;\;S\hat{O}Q = 360 - 108 \\[3ex] Reflex\;\;S\hat{O}Q = 252^\circ \\[3ex] Reflex\;\;S\hat{O}Q = 2 * \angle S\hat{R}Q ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 252 = 2 * S\hat{R}Q \\[3ex] 2 * S\hat{R}Q = 252 \\[3ex] S\hat{R}Q = \dfrac{252}{2} \\[5ex] S\hat{R}Q = 126^\circ \\[3ex] (iii) \\[3ex] O\hat{S}R : O\hat{Q}R = 2:1 \\[3ex] \dfrac{O\hat{S}R}{O\hat{Q}R} = \dfrac{2}{1} \\[5ex] 1 * O\hat{S}R = 2 * O\hat{Q}R \\[3ex] O\hat{S}R = 2 * O\hat{Q}R \\[3ex] Let\;\;O\hat{Q}{R} = m \\[3ex] \implies O\hat{S}{R} = 2m \\[3ex] \underline{Quadrilateral\;SOQR} \\[3ex] \angle S + \angle O + \angle Q + \angle R = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;a\;\;quadrilateral \\[3ex] \angle S = O\hat{S}{R} = 2m \\[3ex] \angle O = Obtuse\;\;S\hat{O}Q = 108^\circ \\[3ex] \angle Q = O\hat{Q}{R} = m \\[3ex] \angle R = S\hat{R}Q = 126^\circ \\[3ex] \implies \\[3ex] 2m + 108 + m + 126 = 360 \\[3ex] 3m + 234 = 360 \\[3ex] 3m = 360 - 234 \\[3ex] 3m = 126 \\[3ex] m = \dfrac{126}{3} \\[5ex] m = 42 \\[3ex] But: \\[3ex] \angle S = O\hat{S}{R} = 2m \\[3ex] \angle O\hat{S}R = 2(42) \\[3ex] \angle O\hat{S}R = 84^\circ $(62.) WASSCE In the diagram, |WX| = |XY| = |YZ| and ∠WXY = 80° What is the size of ∠XWZ? Construction: Draw the isosceles triangles based on the question$ \angle XWY = \angle XYW = \angle YZX = \angle YXZ = k \\[3ex] ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle s\;\;WXY\;\;and\;\;XYZ \\[3ex] \underline{\triangle WXY} \\[3ex] \angle XWY + \angle XYW + \angle WXY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WXY \\[3ex] k + k + 80 = 180 \\[3ex] 2k = 180 - 80 \\[3ex] 2k = 100 \\[3ex] k = \dfrac{100}{2} \\[5ex] k = 50 \\[3ex] \therefore \angle XWY = \angle XYW = \angle YZX = \angle YXZ = 50^\circ \\[3ex] \underline{\triangle XYZ} \\[3ex] \angle YZX + \angle YXZ + \angle XYZ = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XYZ \\[3ex] 50 + 50 + \angle XYZ = 180 \\[3ex] 100 + \angle XYZ = 180 \\[3ex] \angle XYZ = 180 - 100 \\[3ex] \angle XYZ = 80^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;XWZY} \\[3ex] \angle W + \angle Y = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle Y = \angle XYZ = 80^\circ \\[3ex] \angle W = \angle XWZ \\[3ex] \implies \\[3ex] \angle XWZ + 80 = 180 \\[3ex] \angle XWZ = 180 - 80 \\[3ex] \angle XWZ = 100^\circ $(63.) NZQA In this spider web, points A, B, E, and F are all lie on the circumference of a circle, with centre C Lines FE and FA are of equal length Angle AEF = x Angle ADF = y Find the size, w of angle AEB, in terms of x and y Justify your answer with clear geometrical reasoning$ \underline{\triangle AFE} \\[3ex] \angle AEF = \angle EAF = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AFE \\[3ex] \angle AEF + \angle EAF + \angle AFE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AFE \\[3ex] x + x + \angle AFE = 180 \\[3ex] 2x + \angle AFE = 180 \\[3ex] \angle AFE = 180 - 2x \\[3ex] \angle EBD = \angle AFE = 180 - 2x ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;AFEB = interior\;\;opposite\;\;\angle \\[3ex] \angle BED + \angle AEB + \angle AEF = 180 ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle BED + w + x = 180 \\[3ex] \angle BED = 180 - w - x \\[3ex] \angle BDE = \angle ADF = y ...diagram \\[3ex] \underline{\triangle BED} \\[3ex] \angle EBD + \angle BED + \angle BDE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BED \\[3ex] (180 - 2x) + (180 - w - x) + y = 180 \\[3ex] 180 - 2x + 180 - w - x + y = 180 \\[3ex] 180 + 180 - 2x - x + y = 180 + w \\[3ex] 360 - 3x + y = 180 + w \\[3ex] 180 + w = 360 - 3x + y \\[3ex] w = 360 - 3x + y - 180 \\[3ex] w = 180 - 3x + y $(64.) ACT Points A and B lie on the circle below, where central angle$\angle ACB$measures$110^\circ$What is the measure of$\angle ABC$?$ F.\;\; 35^\circ \\[3ex] G.\;\; 40^\circ \\[3ex] H.\;\; 45^\circ \\[3ex] J.\;\; 55^\circ \\[3ex] K.\;\;Cannot\;\;be\;\;determined\;\;from\;\;the\;\;given\;\;information \\[3ex]  \underline{\triangle CAB} \\[3ex] \angle BAC = \angle ABC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle CAB \\[3ex] \angle BAC + \angle ABC + \angle BCA = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CAB \\[3ex] p + p + 110 = 180 \\[3ex] 2p + 110 = 180 \\[3ex] 2p = 180 - 110 \\[3ex] 2p = 70 \\[3ex] p = \dfrac{70}{2} \\[5ex] p = 35 \\[3ex] \therefore \angle ABC = 35^\circ $(65.) NZQA Point C is the centre of the circle. The straight line, SAT, is the tangent to the circle at the point A Prove that the sizes, p, of the angles SAG, AHG, and ABG are all equal to each other. Justify your answer with clear geometrical reasoning.$ \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AG = p \\[3ex] \angle\;\;in\;\;the\;\;alternate\;\;segment\;GHBA = p \\[3ex] Because: \\[3ex] \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AG = \angle\;\;in\;\;the\;\;alternate\;\;segment\;GHBA \\[3ex] p = p \\[3ex] Also: \\[3ex] \angle GHA = \angle GBA = p ...\angle s\;\;in\;\;the\;\;same\;\;major\;\;segment\;GHBA \\[3ex] \therefore p = p = p $(66.) GCSE A, B, C, D, E and F are points on a circle. Circle the line that is a diameter of the circle. BE AD AC BF Let us analyze the options$ If\;\;BE\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle BFE = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But: \\[3ex] \angle BFE + \angle FBE + \angle BEF = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BFE \\[3ex] \angle BFE + 5 + 95 = 180 \\[3ex] \angle BFE + 100 = 180 \\[3ex] \angle BFE = 180 - 100 \\[3ex] \angle BFE = 80^\circ \\[3ex] Because\;\;\angle BFE \ne 90^\circ;\;\;BE \;\;is\;\;NOT\;\;a\;\;diameter \\[3ex] If\;\;AD\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle ACD = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But: \\[3ex] \angle ACD + \angle ADC + \angle CAD = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BFE \\[3ex] \angle ACD + 80 + 10 = 180 \\[3ex] \angle ACD + 90 = 180 \\[3ex] \angle ACD = 180 - 90 \\[3ex] \angle ACD = 90^\circ \\[3ex] Because\;\;\angle ACD = 90^\circ;\;\;AD \;\;is\;\;a\;\;diameter \\[3ex] If\;\;AC\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle ADC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But\;\;\angle ADC = 80^\circ \\[3ex] Because\;\;\angle ADC \ne 90^\circ;\;\;AC \;\;is\;\;NOT\;\;a\;\;diameter \\[3ex] If\;\;BF\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle BEF = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But\;\;\angle BEF = 95^\circ \\[3ex] Because\;\;\angle BEF \ne 90^\circ;\;\;BF \;\;is\;\;NOT\;\;a\;\;diameter $(67.) NSC (67.1) Complete the following theorem: The angle between the tangent to a circle and the chord drawn from the point of contact is equal to ... (67.2) In the diagram below, TVW is the tangent to circle PRVQ at V Chords PV and QR intersect at point S TW || QR$\hat{P} = 42^\circ$Determine, with reasons: (67.2.1) FOUR other angles each equal to 42° (67.2.2) Whether QR is a diameter of the circle (67.2.3) The size of$\hat{Q_2}$is$\hat{V_2} = 67^\circ$(67.1) Complete the following theorem: The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.$ (67.2.1) \\[3ex] First\;\;\angle = 42^\circ: \\[3ex] \angle QRV = 42^\circ ...\angle s\;\;in\;\;the\;\;segment\;\;are\;\;equal \\[3ex] Second\;\;\angle = 42^\circ: \\[3ex] \angle RVW = \hat{V_4} = 42^\circ ... alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;QR\;\;and\;\;TW\;\;are\;\;equal \\[3ex] Third\;\;\angle = 42^\circ: \\[3ex] \angle VQR = \hat{Q_1} = 42^\circ ... \angle \;\;between\;\;tangent\;TVW\;\;and\;\;chord\;VR = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] Fourth\;\;\angle = 42^\circ: \\[3ex] \angle RQV = \hat{V_1} = 42^\circ ... alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;QR\;\;and\;\;TW\;\;are\;\;equal \\[3ex] (67.2.2) \\[3ex] $If QR is a diameter of the circle, then$\angle QVR = 90^\circ...\angle \;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle$If QR is not the diameter of the circle, then$\angle QVR \ne 90^\circ$So, we need to find the value of$\angle QVR \underline{\triangle QVR} \\[3ex] \angle RQV + \angle QRV + \angle QVR = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QVR \\[3ex] 42 + 42 + \angle QVR = 180 \\[3ex] 84 + \angle QVR = 180 \\[3ex] \angle QVR = 180 - 84 \\[3ex] \angle QVR = 96^\circ \\[3ex] Because\;\;\angle QVR \ne 90^\circ: \\[3ex] QR \;\;is\;\;not\;\;a\;\;diameter \\[3ex] (67.2.3) \\[3ex] \underline{\triangle VQP} \\[3ex] \angle QVP + \angle QPV + \angle VQP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle VQP \\[3ex] \angle QVP = \hat{V_2} = 67^\circ \\[3ex] \angle QPV = \hat{P} = 42^\circ \\[3ex] \angle VQP = \hat{Q_1} + \hat{Q_2} \\[3ex] \hat{Q_1} = \angle RQV = 42^\circ \\[3ex] So,\;\; \angle VQP = 42 + \hat{Q_2} \\[3ex] \implies \\[3ex] 67 + 42 + (42 + \hat{Q_2}) = 180 \\[3ex] 109 + 42 + \hat{Q_2} = 180 \\[3ex] 151 + \hat{Q_2} = 180 \\[3ex] \hat{Q_2} = 180 - 151 \\[3ex] \hat{Q_2} = 29^\circ $(68.) ACT In the figure below, A. B, C, and D lie on the circle centered at O Which of the following does NOT appear in the figure? A. Acute triangle B. Equilateral triangle C. Isosceles triangle D. Right triangle E. Scalene triangle Let us analyze the options$ \angle AOB = 60^\circ...implies\;\;an\;\;acute\;\;\triangle \\[3ex] $An acute triangle is a triangle that has three angles less than$90^\circ$Option A appears in the figure$ \underline{\triangle AOB} \\[3ex] OA = OB ...same\;\;radii \\[3ex] \implies \triangle AOB\;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] \angle OBA = \angle OAB ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] $Option C appears in the figure$ \angle OBA + \angle OAB + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] Let\;\;\angle OBA = \angle OAB = p \\[3ex] \angle AOB = 60^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] p + p + 60 = 180 \\[3ex] 2p = 180 - 60 \\[3ex] 2p = 120 \\[3ex] p = \dfrac{120}{2} \\[5ex] p = 60 \\[3ex] \therefore \angle OBA = \angle OAB = 60^\circ \\[3ex] $Each angle in$\triangle AOB = 60^\circ$This implies that$\triangle AOB$is an equilateral triangle. Option B appears in the figure$ \underline{\triangle COD} \\[3ex] \angle COD = 90^\circ ...implies\;\;a\;\;right\;\;\triangle \\[3ex] $A right triangle is a triangle that has one angle of$90^\circ$and two angles less than$90^\circ$Option D appears in the figure The only remaining option is Option E (69.) GCSE A, B and C are points on a circle. CD is a tangent to the circle. Write down the size of angle x Give a reason for your answer.$ x = 65^\circ ...\angle\;\;between\;\;tangent\;CD\;\;and\;\;chord\;CA = \angle\;\;in\;\;the\;\;alternate\;\;segment $(70.) NZQA The points A, B, D, and E all lie on the circumference of a circle, with centre C Angle BDE = x AE = AB Find an expression for angle DBA, in terms of x Justify your answer with clear geometrical reasoning We observe at least two circle theorems with this circle: (1.) Angle in a semicircle: we will need to do some construction to join point B to point E before we can use this theorem (2.) The figure inside the circle is a cyclic quadrilateral. Hence, we shall use at least one of the theorems dealing with cyclic quadrilaterals Construction: Join point B to point E$ \underline{\triangle DEB} \\[3ex] \angle DEB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] Let\;\;\angle DBE = p \\[3ex] \angle BDE + \angle DEB + \angle DBE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DBE \\[3ex] \implies \\[3ex] x + 90 + p = 180 \\[3ex] p = 180 - x - 90 \\[3ex] p = 90 - x \\[3ex] \underline{Cyclic\;\;Quadrilateral\;DEAB} \\[3ex] \angle D + \angle A = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] x + \angle A = 180 \\[3ex] \angle A = 180 - x \\[3ex] \underline{\triangle BEA} \\[3ex] \angle BEA = \angle EBA = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BEA \\[3ex] \angle BAE = \angle A = 180 - x \\[3ex] \angle BEA + \angle EBA + \angle BEA = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BEA \\[3ex] \implies \\[3ex] k + k + (180 - x) = 180 \\[3ex] 2k + 180 - x = 180 \\[3ex] 2k = 180 - 180 + x \\[3ex] 2k = x \\[3ex] k = \dfrac{x}{2} \\[5ex] \angle DBA = p + k ...diagram \\[3ex] \angle DBA = (90 - x) + \dfrac{x}{2} \\[5ex] \angle DBA = 90 - x + \dfrac{x}{2} \\[5ex] \angle DBA = \dfrac{90}{1} - \dfrac{x}{1} + \dfrac{x}{2} \\[5ex] \angle DBA = \dfrac{180 - 2x + x}{2} \\[5ex] \angle DBA = \dfrac{180 - x}{2} $(71.) WASSCE (a.) PQ is a tangent to a curve RST at the point S PRT is a straight line,$\angle TPS = 34^\circ$and$\angle TSQ = 65^\circ$(I.) Illustrate the information in a diagram (II.) Find the value of: (i.)$\angle RTS$(ii.)$\angle SRP$(b.) In the diagram, |VZ| = |YZ|,$\angle YXZ = 20^\circ$and$\angle ZVY = 52^\circ$Calculate the size of$\angle WYZ$(a.) (I.) Based on the information, the diagram is drawn as shown:$ (II.) \\[3ex] (i.) \\[3ex] \underline{\triangle PTS} \\[3ex] \angle PST = \angle RTS + 34 ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;\angle s \\[3ex] 65 = \angle RTS + 34 \\[3ex] \angle RTS + 34 = 65 \\[3ex] \angle RTS = 65 - 34 \\[3ex] \angle RTS = 31^\circ \\[3ex] (ii.) \\[3ex] $Construction: Join point S to point R$ \angle TSQ = \angle SRT = 65^\circ ...\angle \;\;between\;\;tangent\;PSQ\;\;and\;\;chord\;ST = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle SRP + \angle SRT = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle SRP + 65 = 180 \\[3ex] \angle SRP = 180 - 65 \\[3ex] \angle SRP = 115^\circ \\[3ex] (b.) \\[3ex]  \angle ZVY = \angle ZYV = 52^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ZVY \\[3ex] \underline{\triangle XYZ} \\[3ex] \angle YXZ + \angle ZYX + \angle XZY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XYZ \\[3ex] \angle ZYX = \angle ZYV = 52^\circ ...diagram \\[3ex] 20 + 52 + \angle XZY = 180 \\[3ex] 72 + \angle XZY = 180 \\[3ex] \angle XZY = 180 - 72 \\[3ex] \angle XZY = 108^\circ \\[3ex] \angle YWZ = \angle ZVY = 52^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle WYZ} \\[3ex] \angle WYZ + \angle YWZ + \angle WZY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WYZ \\[3ex] \angle WZY = \angle XZY = 108^\circ...diagram \\[3ex] \implies \\[3ex] \angle WYZ + 52 + 108 = 180 \\[3ex] \angle WYZ + 160 = 180 \\[3ex] \angle WYZ = 180 - 160 \\[3ex] \angle WYZ = 20^\circ $(72.) CSEC (a.) The circle shown below has centre O and the points A, B, C and D lying on the circumference. A straight line passes through the points A and B Angle CBD = 49° and angle OAB = 37° (i) Write down the mathematical names of the straight lines BC and OA BC ....................................................... OA ....................................................... (ii) Determine the value of EACH of the following angles. Show detailed working where necessary and give a reason to support your answer. (a) x (b) y (i) BC is a chord OA is a radius$ (ii) \\[3ex] (a) \\[3ex] \angle BAO = \angle ABO = 37^\circ ...base\;\;angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \underline{\triangle AOB} \\[3ex] \angle AOB + \angle BAO + \angle ABO = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB = x...diagram \\[3ex] \implies \\[3ex] x + 37 + 37 = 180 \\[3ex] x + 74 = 180 \\[3ex] x = 180 - 74 \\[3ex] x = 106^\circ \\[3ex] (b) \\[3ex] \underline{\triangle BCD} \\[3ex] \angle BCD = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle BDC + \angle BCD + \angle CBD = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BCD \\[3ex] \angle BDC = y...diagram \\[3ex] \implies \\[3ex] y + 90 + 49 = 180 \\[3ex] y + 139 = 180 \\[3ex] y = 180 - 139 \\[3ex] y = 41^\circ \\[3ex] $Common Mistake:$y = 37^\circ...\angle s\;\;in\;\;the\;\;segment$No$37^\circ$and$y^\circ$are NOT in the same segment because |OA| did not extend to point C. (There is no |AC|) (73.) USSCE Advance Mathematics Paper 2 (a.) Fill in the missing words. The angle between a tangent and a chord through the point of contact is equal to the angle$\rule{4cm}{0.15mm}$by the chord in the alternate segment. (b.) Two circles intersect at points P and Q Line APB is drawn through P The tangents at A and B meet at C If$\angle ABC = \beta^\circ$and$\angle BAC = \alpha^\circ$, find an expression in terms of$\alpha$and$\beta$for$ (i.)\;\; \angle PQB \\[3ex] (ii.)\;\; \angle PQA \\[3ex] (iii.)\;\; \angle AQB \\[3ex] $(a.) The angle between a tangent and a chord through the point of contact is equal to the angle$\underline{subtended}$by the chord in the alternate segment.$ (b.) \\[3ex] (i.) \\[3ex] \angle \;\;between\;\;tangent\;BC\;\;and\;\;chord\;BP = \beta \\[3ex] \angle \;\;in\;\;the\;\;alternate\;\;segment = \angle PQB \\[3ex] \beta = \angle PQB... \angle \;\;between\;\;tangent\;BC\;\;and\;\;chord\;BP = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle PQB = \beta^\circ \\[3ex] (ii.) \\[3ex] \angle \;\;between\;\;tangent\;AC\;\;and\;\;chord\;AP = \alpha \\[3ex] \angle \;\;in\;\;the\;\;alternate\;\;segment = \angle PQA \\[3ex] \alpha = \angle PQA... \angle \;\;between\;\;tangent\;AC\;\;and\;\;chord\;AP = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle PQA = \alpha^\circ \\[3ex] (iii.) \\[3ex] \angle AQB = \angle PQA + \angle PQB \\[3ex] \angle AQB = \alpha + \beta $(74.) ACT In the figure below, lines l and m are tangent to the circle at points B and D, respectively. Points A and C are on the circle. The measure of$\angle ABC$is$95^\circ$and the measure of$\angle BCD$is$85^\circ$The lines in which of the following pairs of lines are necessarily parallel?$ I.\;\; l\;\;and\;\;m \\[3ex] II.\;\; \overleftrightarrow{AB} \;\;and\;\; \overleftrightarrow{DC} \\[3ex] III.\;\; \overleftrightarrow{AD} \;\;and\;\; \overleftrightarrow{BC} \\[3ex] F.\;\; I\;\;only \\[3ex] G.\;\; II\;\;only \\[3ex] H.\;\; III\;\;only \\[3ex] J.\;\; I\;\;and\;\;II\;\;only \\[3ex] K.\;\; I,\;\;II,\;\;and\;\;III \\[3ex]  l\;\;and\;\;m\;\;are\;\;not\;\;parallel \\[3ex] Eliminate\;\;Option\;I \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle A + \angle C = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A + 85 = 180 \\[3ex] \angle A = 180 - 85 \\[3ex] \angle A = 95^\circ \\[3ex] Also: \\[3ex] \angle B + \angle D = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 95 + \angle D = 180 \\[3ex] \angle D = 180 - 95 \\[3ex] \angle D = 85^\circ \\[3ex] Try\;\;Option\;II \\[3ex]  If\;\;\overleftrightarrow{AB} \;||\; \overleftrightarrow{DC} \\[3ex] e = 85^\circ ...corresponding\;\;\angle s\;\;are\;\;congruent \\[3ex] But: \\[3ex] e + 95^\circ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] e = 180 - 95 \\[3ex] e = 85^\circ \\[3ex] 85^\circ = 85^\circ \\[3ex] \implies that\;\; \overleftrightarrow{AB} \;||\; \overleftrightarrow{DC} \\[3ex] Try\;\;Option\;III \\[3ex]  If\;\;\overleftrightarrow{AD} \;||\; \overleftrightarrow{BC} \\[3ex] f = 95^\circ ...corresponding\;\;\angle s\;\;are\;\;congruent \\[3ex] But: \\[3ex] f + 95^\circ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] f = 180 - 95 \\[3ex] f = 85^\circ \\[3ex] 95^\circ \ne 85^\circ \\[3ex] \implies that\;\; \overleftrightarrow{AD} \;\not||\; \overleftrightarrow{BC} \\[3ex] Eliminate\;\;Option\;III \\[3ex] Correct\;\;Option:\;\;II\;\;only $(75.) NZQA In the diagram below, straight line SRT is the tangent to the circle at the point R The triangle UVR is isosceles, with UV = RV Angle UVR = 38° Calculate the size, q, of angle VRS Justify your answer with clear geometrical reasoning.$ q = \angle VUR ...\angle \;\;between\;\;tangent\;SRT \;\;and\;\;chord\;VR = \angle\;\;in\;\;the\;\;second\;\;segment \\[3ex] \underline{\triangle UVR} \\[3ex] \angle VUR = \angle VRU = q...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \implies \\[3ex] \angle VUR + \angle VRU + \angle UVR = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle UVR \\[3ex] q + q + 38 = 180 \\[3ex] 2q + 38 = 180 \\[3ex] 2q = 180 - 38 \\[3ex] 2q = 142 \\[3ex] q = \dfrac{142}{2} \\[5ex] q = 71^\circ $(76.) ACT In the figure below, the circle with center O has a radius of 7 inches and the measure of$\overset{\huge\frown}{AB}$is 80° What is the measure of$\angle BAO$?$ F.\;\; 30^\circ \\[3ex] G.\;\; 40^\circ \\[3ex] H.\;\; 50^\circ \\[3ex] J.\;\; 60^\circ \\[3ex] K.\;\; 80^\circ \\[3ex] $Construction: Join the radius from point O to point B Intercepted arc = angle at center (central angle)$ \angle AOB = \overset{\huge\frown}{AB} = 80^\circ \\[3ex] \underline{\triangle OAB} \\[3ex] \angle BAO = \angle ABO = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OAB \\[3ex] \angle BAO + \angle ABO + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAB \\[3ex] \implies \\[3ex] p + p + 80 = 180 \\[3ex] 2p = 180 - 80 \\[3ex] 2p = 100 \\[3ex] p = \dfrac{100}{2} \\[5ex] p = 50 \\[3ex] \therefore \angle BAO = \angle ABO = 50^\circ $(77.) curriculum.gov.mt (a) (i) Explain why a = 90 (ii) Work out the size of angles b and c (b) (i) AB and CD are two chords of a circle intersecting at E. Show that triangles ADE and CBE are similar. (ii) AD : CB = 3 : 2 and BC = 4 cm. Work out the length of AD.$ (a) \\[3ex] (i) \\[3ex] a = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex]  (ii) \\[3ex] b^\circ = c^\circ ...\angle\;\;between\;\;tangent\;EG\;\;and\;\;chord\;EF = \angle\;\;in\;\;alternate\;\;segment \\[3ex] Also: \\[3ex] \angle EFG = c^\circ ...\angle\;\;between\;\;tangent\;FG\;\;and\;\;chord\;EF = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle EFG} \\[3ex] \angle FEG + \angle EFG + \angle FGE = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle EFG \\[3ex] b + c + 52 = 180 \\[3ex] But\;\; b = c \\[3ex] c + c + 52 = 180 \\[3ex] 2c = 180 - 52 \\[3ex] 2c = 128 \\[3ex] c = \dfrac{128}{2} \\[5ex] c = 64^\circ \\[3ex] b = c = 64^\circ \\[5ex] (b) \\[3ex]  (i) \\[3ex] If:\;\;\triangle ADE \sim \triangle CBE \\[3ex] Then:\;\; \dfrac{AE}{CE} = \dfrac{ED}{EB} ...Similarity\;\;Ratio \;(Scale\;\;Factor) \\[5ex] AE * EB = CE * ED ... Intersecting\;\;Chords\;\;Theorem \\[3ex] \implies \\[3ex] \dfrac{AE}{CE} = \dfrac{ED}{EB} \\[5ex] (ii) \\[3ex] AD : CB = 3 : 2 \\[3ex] AD : BC = 3 : 2 \\[3ex] \implies \\[3ex] \dfrac{AD}{BC} = \dfrac{3}{2} \\[5ex] BC = 4\;cm \\[3ex] AD = ? \\[3ex] \implies \\[3ex] \dfrac{AD}{BC} = \dfrac{3}{2} = \dfrac{AD}{4} \\[5ex] \dfrac{AD}{4} = \dfrac{3}{2} \\[5ex] AD = \dfrac{3 * 4}{2} \\[5ex] AD = 3(2) \\[3ex] AD = 6\;cm $(78.) GCSE A, B, C, D and E are points on a circle. BFD and AFC are straight lines. DC = DF Work out the size of angle x You must show your working which may be on the diagram. Construction: Join the chord from point B to point C$ \angle BDC = \angle BAC = 24^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle AFE + \angle EFD + \angle DFC = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] x + 2x + \angle DFC = 180 \\[3ex] 3x + \angle DFC = 180 \\[3ex] \angle DFC = 180 - 3x \\[3ex] \underline{\triangle DFC} \\[3ex] \angle DFC = \angle DCF = 180 - 3x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DFC \\[3ex] \angle DFC + \angle DCF + \angle FDC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DFC \\[3ex] \angle FDC = \angle BDC = 24^\circ ...diagram \\[3ex] \implies \\[3ex] (180 - 3x) + (180 - 3x) + 24 = 180 \\[3ex] 180 - 3x + 180 - 3x + 24 = 180 \\[3ex] 384 - 6x = 180 \\[3ex] 384 - 180 = 6x \\[3ex] 204 = 6x \\[3ex] 6x = 204 \\[3ex] x = \dfrac{204}{6} \\[5ex] x = 34^\circ $(79.) CSEC The diagram below, not drawn to scale, shows a circle. The points P, Q, R, T and V are on the circumference. QRS is a straight line. Angle PVR = 75° and angle TRS = 60° Determine the value of EACH of the following angles. Show detailed working where necessary and give a reason to support your answer. (i) Angle PTR (ii) Angle TPQ (iii) Obtuse angle POR where O is the centre of the circle.$ (i) \\[3ex] \angle PTR = 75^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PTRQ} \\[3ex]  \angle TPQ = 60^\circ ...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;PTRQ = interior\;\;opposite\;\;\angle \\[3ex] $Construction: (a.) Indicate the centre of the circle, O (b.) Draw the line from point P to centre O (c.) Draw the line from centre O to point R$ Obtuse\;\;\angle POR = 2 * \angle PVR ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle POR = 2 * 75 \\[3ex] \angle POR = 150^\circ $(80.) ACT In the figure below, line q is parallel to line r, C is the center of the circle.$\overline{AE}$and$\overline{BD}$both go through C, q is tangent to the circle at B, F lies on$\overleftrightarrow{AD}$, and the measure of$\angle FAC$is$145^\circ$What is the measure of$\angle BCE$?$ F.\;\; 145^\circ \\[3ex] G.\;\; 90^\circ \\[3ex] H.\;\; 65^\circ \\[3ex] J.\;\; 55^\circ \\[3ex] K.\;\; 35^\circ \\[3ex]  \theta = 145^\circ...interior\;\;alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BEC + \theta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle BEC + 145 = 180 \\[3ex] \angle BEC = 180 - 145 \\[3ex] \angle BEC = 35^\circ \\[3ex] \angle CBE = 90^\circ \\[3ex] ...radius\;CB\;\perp\;tangent\;qBE\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \underline{\triangle CBE} \\[3ex] \angle BCE + \angle BEC + \angle CBE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CBE \\[3ex] \angle BCE + 35 + 90 = 180 \\[3ex] \angle BCE + 125 = 180 \\[3ex] \angle BCE = 180 - 125 \\[3ex] \angle BCE = 55^\circ $(81.) CSEC The diagram below, not drawn to scale, shows a circle, centre O The lines BD and DCE are tangents to the circle, and angle BCD = 79° Calculate, giving reasons for your answer, the size of EACH of the following angles:$ (i)\:\: \angle OCE \\[3ex] (ii)\:\: \angle BAC \\[3ex] (iii)\:\: \angle BOC \\[3ex] (iv)\:\: \angle BDC \\[3ex]  (i) \\[3ex] \angle OCE = 90^\circ ...radius\;OC \; \perp\; tangent\;DCE \;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] (ii) \\[3ex] \angle BCD = \angle BAC = 70^\circ ... \angle\;\;between\;\;chord\;BC\;\;and\;\;tangent\;DCE = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] (iii) \\[3ex] \angle BOC = 2 * \angle BAC ...\angle \;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * 70 \\[3ex] \angle BOC = 140^\circ \\[3ex] (iv) \\[3ex] |BD| = |DCE| ...two\;\;tangents:\;BD\;\;and\;\;DCE\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;A\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \underline{\triangle BDC} \\[3ex] \angle BCD = \angle CBD = 70^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BDC \\[3ex] \angle BCD + \angle CBD + \angle BDC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BDC \\[3ex] 70 + 70 + \angle BDC = 180 \\[3ex] 140 + \angle BDC = 180 \\[3ex] \angle BDC = 180 - 140 \\[3ex] \angle BDC = 40^\circ $(82.) GCSE A and B are points on a circle, centre O. BC is a tangent to the circle. AOC is a straight line. Angle ABO = x° Find the size of angle ACB, in terms of x Gve your answer in its simplest form. Give reasons for each stage of your working. Construction: Join the radius from the centre, O to point B on the circumference$ \angle ABO = \angle BAO = x ...base\;\;\angle s \;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle OBC = 90^\circ \\[3ex] ...radius\;OB\;\perp\;tangent\;BC\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \angle BOC = x + x \\[3ex] ... exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle BOC = 2x \\[3ex] \underline{\triangle OBC} \\[3ex] \angle BOC + \angle OBC + \angle OCB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OBC \\[3ex] 2x + 90 + \angle OCB = 180 \\[3ex] 2x + 90 + \angle OCB = 180 \\[3ex] \angle OCB = 180 - 2x - 90 \\[3ex] \angle OCB = 90 - 2x \\[3ex] \angle OCB = 2(45 - x) \\[3ex] \angle OCB = \angle ACB ...diagram \\[3ex] \therefore \angle ACB = 2(45 - x) $(83.) GCSE A, B and C are points on a circle. DCB is a straight line. PAQ is a tangent to the circle. Sam is trying to work out the size of angle m. Here is his working. Make a criticism of his working. Sam got the answer correctly. However, the reasons are incorrect. He is supposed to state his working as:$ \angle ACB = 56^\circ...\angle \;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AB = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] m = 180 - 56...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] m = 124^\circ $(84.) ACT In the circle shown below, M is the center and lies on$\overline{RU}$and$\overline{ST}$Which of the following statements is NOT true?$ F.\;\; \angle TUM \;\;measures\;\;65^\circ \\[3ex] G.\;\; \overline{TU}\;\;is\;\;parallel\;\;to\;\;\overline{RS} \\[3ex] H.\;\; \overset{\Huge\frown}{TXU}\;\;measures\;\;50^\circ \\[3ex] J.\;\; \overline{RM} \cong \overline{TM} \\[3ex] K.\;\; \overline{RS} \cong \overline{SM} \\[3ex] $Let us analyze the options$ \angle TMU = 50^\circ ...vertical \angle s\;\;are\;\;equal \\[3ex] \underline{\triangle MTU} \\[3ex] \angle UTM = \angle TUM = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MTU \\[3ex] \angle UTM + \angle TUM + \angle TMU = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MTU \\[3ex] p + p + 50 = 180 \\[3ex] 2p + 50 = 180 \\[3ex] 2p = 180 - 50 \\[3ex] 2p = 130 \\[3ex] p = \dfrac{130}{2} \\[5ex] p = 65 \\[3ex] \therefore \angle UTM = \angle TUM = 65^\circ \\[3ex] Option\;\;F\;\;is\;\;true \\[3ex] \underline{\triangle MRS} \\[3ex] \angle SRM = \angle RSM = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MRS \\[3ex] \angle SRM + \angle RSM + \angle RMS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MRS \\[3ex] k + k + 50 = 180 \\[3ex] 2k + 50 = 180 \\[3ex] 2k = 180 - 50 \\[3ex] 2k = 130 \\[3ex] k = \dfrac{130}{2} \\[5ex] k = 65 \\[3ex] \therefore \angle SRM = \angle RSM = 65^\circ \\[3ex] \angle SRM = \angle SRU = 65^\circ ...diagram \\[3ex] \angle TUM = \angle TUR = 65^\circ ...diagram \\[3ex] Because: \\[3ex] \angle SRU = \angle TUR = 65^\circ ...alternate \angle s\;\;are\;\;equal \\[3ex] \implies \;\;that\;\; \overline{RS} \;\;||\;\; \overline{TU} \\[3ex] Option\;\;G\;\;is\;\;true \\[3ex] \overset{\Huge\frown}{TXU} = 50^\circ ...vertical \angle s\;\;are\;\;equal \\[3ex] ...also\;\;\angle\;\;subtended\;\;by\;\;arc\;TXU \\[3ex] Option\;\;H\;\;is\;\;true \\[3ex] \overline{RM} = \overline{TM} ...same\;\;radii\;\;from\;\;center\;M \\[3ex] Option\;\;J\;\;is\;\;true \\[3ex] For\;\;the\;\;remaining\;\; Option\;\;K \\[3ex] \overline{RM} = \overline{SM} ...same\;\;radii\;\;from\;\;center\;M \\[3ex] \implies\;\; \triangle RMS\;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] However: \\[3ex] \overline{RS} \ncong \overline{SM} \\[3ex] $Because$\triangle RMS$is not an equilateral triangle An equilateral triangle must have each of the three angles as$60^\circ$in order for all sides to be congruent$ Option\;\;K\;\;is\;\;NOT\;\;true $(85.) NZQA (a) A climbing frame is made from a semi-circle and triangles. The climbing frame is symmetrical about FE. Angle CAD = 33° AD is the diameter of the semi-circle. (i) Calculate the size, x, of the angle ALD Justify your answer with clear geometric reasoning. (ii) Calculate the size, y, of angle BAC Justify your answer with clear geometric reasoning.$ (i) \\[3ex] \angle LAD = \angle CAD = 33^\circ...diagram \\[3ex] \underline{\triangle ALD} \\[3ex] Because\;\;the\;\;climbing\;\;frame\;\;is\;\;symmetrical\;\;about\;\;FE: \\[3ex] \angle LDA = \angle LAD = 33^\circ \\[3ex] \angle LDA + \angle LAD + \angle ALD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ALD \\[3ex] 33 + 33 + x = 180 \\[3ex] 66 + x = 180 \\[3ex] x = 180 - 66 \\[3ex] x = 114^\circ \\[3ex] (ii) \\[3ex] \angle DBA = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle ALD = \angle DBA + \angle BAC ... exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] x = 90 + y \\[3ex] 90 + y = x \\[3ex] y = x - 90 \\[3ex] y = 114 - 90 \\[3ex] y = 24^\circ $(86.) CSEC In the diagram below, not drawn to scale, O is the centre of the circle. The lines SK and AF are parallel.$\angle KSW = 62^\circ\angle SAF = 54^\circ$Calculate, giving reasons for your answer, the measure of: (i)$\angle FAW$(ii)$\angle SKF$(iii)$\angle ASW (i) \\[3ex] \angle SAW = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle SAF + \angle FAW = \angle SAW ...diagram \\[3ex] 54 + \angle FAW = 90 \\[3ex] \angle FAW = 90 - 54 \\[3ex] \angle FAW = 36^\circ \\[3ex] (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;SKFA} \\[3ex] \angle A + \angle K = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A = \angle SAF = 54^\circ \\[3ex] \angle K = \angle SKF \\[3ex] \implies \\[3ex] 54 + \angle SKF = 180 \\[3ex] \angle SKF = 180 - 54 \\[3ex] \angle SKF = 126^\circ \\[3ex] (iii) \\[3ex]  \theta = 62^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle ASB + 54 + \theta = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SBA \\[3ex] \angle ASB + 54 + 62 = 180 \\[3ex] \angle ASB + 116 = 180 \\[3ex] \angle ASB = 180 - 116 \\[3ex] \angle ASB = 64^\circ \\[3ex] \angle ASW = \angle ASB ...diagram \\[3ex] \therefore \angle ASW = 64^\circ $(87.) WASSCE In the diagram,$\overline{MN}$and$\overline{MQ}$are tangents to the circle centre O If$\angle MNQ = x$,$\angle NMQ = y$and$\angle NQP = 46^\circ$, find the value of: (i) x (ii) y Construction: Join the chord from point N to point P$ (i) \\[3ex] \angle QNP = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] x = \angle QPN ...\angle\;\;between\;\;tangent\;MN\;\;and\;\;a\;\;chord\;NQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle NQP} \\[3ex] \angle QNP + \angle QPN + \angle NQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle NQP \\[3ex] 90 + x + 46 = 180 \\[3ex] x + 136 = 180 \\[3ex] x = 180 - 136 \\[3ex] x = 44^\circ \\[3ex] (ii) \\[3ex] \angle PQM = 90^\circ ...radius\;OQ\;\perp\;tangent\;MQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle NQM + \angle NQP = \angle PQM ...diagram \\[3ex] \angle NQM + 46 = 90 \\[3ex] \angle NQM = 90 - 46 \\[3ex] \angle NQM = 44^\circ \\[3ex] \underline{\triangle MNQ} \\[3ex] \angle MNQ + \angle NMQ + \angle NQM = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNQ \\[3ex] x + y + 44 = 180 \\[3ex] 44 + y + 44 = 180 \\[3ex] y + 88 = 180 \\[3ex] y = 180 - 88 \\[3ex] y = 92^\circ \\[3ex] OR \\[3ex] (i) \;\;and\;\; (ii) \\[3ex] \angle PQM = 90^\circ ...radius\;OQ\;\perp\;tangent\;MQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle NQM + \angle NQP = \angle PQM ...diagram \\[3ex] \angle NQM + 46 = 90 \\[3ex] \angle NQM = 90 - 46 \\[3ex] \angle NQM = 44^\circ \\[3ex] |MN| = |MQ|...two\;\;tangents\;\;MN\;\;and\;\;MQ\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point\;M\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \angle NQM = \angle MNQ = 44^\circ = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MNQ \\[3ex] \underline{\triangle MNQ} \\[3ex] \angle MNQ + \angle NMQ + \angle NQM = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNQ \\[3ex] 44 + y + 44 = 180 \\[3ex] y + 88 = 180 \\[3ex] y = 180 - 88 \\[3ex] y = 92^\circ $(88.) ACT In the circle shown below, chords$\overline{TR}$and$\overline{QS}$intersect at P, which is the center of the circle, and the measure of$\angle PST$is$30^\circ$What is the degree measure of minor arc$\overset{\huge\frown}{RS}$?$ F.\;\; 30^\circ \\[3ex] G.\;\; 45^\circ \\[3ex] H.\;\; 60^\circ \\[3ex] J.\;\; 90^\circ \\[3ex] K.\;\; Cannot\;\;be\;\;determined\;\;from\;\;the\;\;given\;\;information \\[3ex]  First\;\;Approach:\;\;Recommmended\;\;for\;\;ACT \\[3ex] \underline{\triangle PST} \\[3ex] \angle PST = \angle PTS = 30^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PST \\[3ex] \overset{\huge\frown}{RS} = \angle PST + \angle PTS \\[3ex] ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s\\[3ex] \overset{\huge\frown}{RS} = 30 + 30 \\[3ex] \overset{\huge\frown}{RS} = 60^\circ \\[3ex] OR \\[3ex] Second\;\;Approach: \\[3ex] \angle PST + \angle PTS + \angle TPS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PST \\[3ex] 30 + 30 + \angle TPS = 180 \\[3ex] 60 + \angle TPS = 180 \\[3ex] \angle TPS = 180 - 60 \\[3ex] \angle TPS = 120^\circ \\[3ex] \angle TPS + \overset{\huge\frown}{RS} = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 120 + \overset{\huge\frown}{RS} = 180 \\[3ex] \overset{\huge\frown}{RS} = 180 - 120 \\[3ex] \overset{\huge\frown}{RS} = 60^\circ $(89.) NZQA A circular hoop is hung with wires running through it. O is the centre of the circular hoop. Angle OAB = 50° Angle ODB = 22° Calculate the size, z, of angle EJO Justify your answer with clear geometric reasoning.$ \underline{\triangle AOD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] \angle OAD = \angle OAB = 50^\circ \\[3ex] \angle ODA = \angle ODB = 22^\circ \\[3ex] \implies \\[3ex] 50 + 22 + \angle AOD = 180 \\[3ex] 72 + \angle AOD = 180 \\[3ex] \angle AOD = 180 - 72 \\[3ex] \angle AOD = 108^\circ \\[3ex] \angle AOD = \angle EOF = 108^\circ ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle EOF} \\[3ex] \angle OEF = \angle OFE = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle EOF \\[3ex] \angle OEF + \angle OFE + \angle EOF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOF \\[3ex] p + p + 108 = 180 \\[3ex] 2p = 180 - 108 \\[3ex] 2p = 72 \\[3ex] p = \dfrac{72}{2} \\[5ex] p = 36 \\[3ex] \therefore \angle OEF = \angle OFE = 36^\circ \\[3ex] Also: \\[3ex] \angle OBD = 90^\circ ... radius\;OB \;\perp\;tangent\;ABD\;\;at\;\;a\;\;point\;\;of\;\;contact\;B \\[3ex] \underline{\triangle BOD} \\[3ex] \angle BOD + \angle OBD + \angle ODB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOD \\[3ex] \angle BOD + 90 + 22 = 180 \\[3ex] \angle BOD + 112 = 180 \\[3ex] \angle BOD = 180 - 112 \\[3ex] \angle BOD = 68^\circ \\[3ex] \angle EOJ = \angle BOD = 68^\circ ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle EOJ} \\[3ex] \angle EOJ + \angle OEJ + \angle EJO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOJ \\[3ex] \angle OEJ = \angle OEF = 36^\circ ...diagram \\[3ex] \implies \\[3ex] 68 + 36 + z = 180 \\[3ex] 104 + z = 180 \\[3ex] z = 180 - 104 \\[3ex] z = 76^\circ $(90.) WASSCE In the diagram, PQRST is a circle with centre O RS || QT, |QR| = |RS| and ∠QTS = 52° Find: (i) ∠SQT (ii) ∠PQT$ \underline{Cyclic\;\;Quadrilateral\;QRST} \\[3ex] \angle R + \angle T = 180^\circ \\[3ex] ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle R + 52 = 180 \\[3ex] \angle R = 180 - 52 \\[3ex] \angle R = 128^\circ \\[3ex] \underline{\triangle QRS} \\[3ex] \angle RQS = \angle RSQ = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle QRS \\[3ex] \angle RQS + \angle RSQ + \angle QRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QRS \\[3ex] \angle QRS = \angle R = 128^\circ \\[3ex] k + k + 128 = 180 \\[3ex] 2k = 180 - 128 \\[3ex] 2k = 52 \\[3ex] k = \dfrac{52}{2} \\[5ex] k = 26 \\[3ex] \therefore \angle RQS = \angle RSQ = 26^\circ \\[3ex] (i) \\[3ex] \angle RSQ = \angle SQT = 26^\circ ...alternate \;\;\angle s\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \angle PQS = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle PQS = \angle PQT + \angle SQT ...diagram \\[3ex] 90 = \angle PQT + 26 \\[3ex] \angle PQT + 26 = 90 \\[3ex] \angle PQT = 90 - 26 \\[3ex] \angle PQT = 64^\circ $(91.) GCSE A, B, C and D are points on the circumference of a circle, centre O. FDE is a tangent to the circle. (a) Show that$y - x = 90$You must give a reason for each stage of your working. Dylan was asked to give some possible values for x and y. He said, "y could be 200 and x could be 110, because 200 - 110 = 90" (b) Is Dylan correct? You must give a reason for your answer. Construction: Join the radius from the centre O to the point B on the circumference$ (a) \\[3ex] \angle ODB = \angle OBD = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOD \\[3ex] \underline{\triangle BOD} \\[3ex] \angle ODB + Obtuse\;\angle OBD + \angle BOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOD \\[3ex] x + x + Obtuse\;\angle BOD = 180 \\[3ex] 2x + Obtuse\;\angle BOD = 180 \\[3ex] Obtuse\;\angle BOD = 180 - 2x \\[3ex] Obtuse\;\angle BOD + Reflex\;\angle BOD = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] (180 - 2x) + Reflex\;\angle BOD = 360 \\[3ex] Reflex\;\angle BOD = 360 - (180 - 2x) \\[3ex] Reflex\;\angle BOD = 360 - 180 + 2x \\[3ex] Reflex\;\angle BOD = 180 + 2x \\[3ex] Reflex\;\angle BOD = 2 * \angle BAD ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 180 + 2x = 2 * y \\[3ex] 180 + 2x = 2y \\[3ex] 2y = 180 + 2x \\[3ex] 2y = 2(90 + x) \\[3ex] Divide\;\;both\;\;sides\;\;by\;\;2 \\[3ex] y = 90 + x \\[3ex] \therefore y - x = 90 \\[3ex] (b) \\[3ex] Based\;\;on\;\;Dylan's\;\;response: \\[3ex] If\;\;y = 200: \\[3ex] Then\;\; Reflex\;\;\angle BOD = 2y = 2(200) = 400 \\[3ex] 400\;\;is\;\;not\;\;a\;\;Reflex\;\;\angle \\[3ex] This\;\;is\;\;not\;\;possible \\[3ex] Dylan\;\;is\;\;not\;\;correct $(92.) ACT The measure of any angle inscribed in a circle is$\dfrac{1}{2}$the measure of the intercepted arc. Points A, B, and C are on the circle shown below. What is the measure of ∠ABC?$ A.\;\; 40^\circ \\[3ex] B.\;\; 60^\circ \\[3ex] C.\;\; 80^\circ \\[3ex] D.\;\; 140^\circ \\[3ex] E.\;\; 160^\circ \\[3ex] $Construction: Join the chord from point A to point C$ 190 = 2 * \angle BAC ...intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] 2 * \angle BAC = 190 \\[3ex] \angle BAC = \dfrac{190}{2} \\[5ex] \angle BAC = 95^\circ \\[3ex] 90 = 2 * \angle BCA ...intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] 2 * \angle BCA = 90 \\[3ex] \angle BCA = \dfrac{90}{2} \\[5ex] \angle BCA = 45^\circ \\[3ex] \underline{\triangle BAC} \\[3ex] \angle ABC + \angle BAC + \angle BCA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BAC \\[3ex] \angle ABC + 95 + 45 = 180 \\[3ex] \angle ABC + 140 = 180 \\[3ex] \angle ABC = 180 - 140 \\[3ex] \angle ABC = 40^\circ $(93.) CSEC In the diagram below, not drawn to scale, PQ is a tangent to the circle, centre O PR is parallel to OS and angle SPR = 26° Calculate, giving reasons for your answer, the size of (i) angle PTS (ii) angle RPQ$ (i) \\[3ex] \angle OSP = 26^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OSP = \angle OPS = 26^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POS \\[3ex] \angle OSP + \angle OPS + \angle POS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POS \\[3ex] 26 + 26 + \angle POS = 180 \\[3ex] 52 + \angle POS = 180 \\[3ex] \angle POS = 180 - 52 \\[3ex] \angle POS = 128^\circ \\[3ex] \angle POS = 2 * \angle PTS ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 128 = 2 * \angle PTS \\[3ex] 2 * \angle PTS = 128 \\[3ex] \angle PTS = \dfrac{128}{2} \\[5ex] \angle PTS = 64^\circ \\[3ex] (ii) \\[3ex] \angle SPQ = \angle PTS = 64^\circ... \angle\;\;between\;\;tangent\;PQ\;\;and\;\;chord\;PS = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle SPQ = \angle SPR + \angle RPQ ...diagram \\[3ex] 64 = 26 + \angle RPQ \\[3ex] 26 + \angle RPQ = 64 \\[3ex] \angle RPQ = 64 - 26 \\[3ex] \angle RPQ = 38^\circ $(94.) WASSCE The diagram shows a circle PQRS with centre O$\angle UQR = 68^\circ$,$\angle TPS = 74^\circ$,$\angle QSR = 40^\circ$Calculate the value of$\angle PRS \angle UQR = \angle PSR = 68^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quad\;PQRS = interior\;\;opposite\;\;\angle \\[3ex] \angle QSR = \angle RPQ = 40^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle TPS + \angle SPR + \angle RPQ = 180^\circ \\[3ex] ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 74 + \angle SPR + 40 = 180 \\[3ex] 114 + \angle SPR = 180 \\[3ex] \angle SPR = 180 - 114 \\[3ex] \angle SPR = 66^\circ \\[3ex] \underline{\triangle PSR} \\[3ex] \angle PSR + \angle PRS + \angle SPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] 68 + \angle PRS + 66 = 180 \\[3ex] 134 + \angle PRS = 180 \\[3ex] \angle PRS = 180 - 134 \\[3ex] \angle PRS = 46^\circ \\[3ex] OR \\[3ex] \angle TPS = \angle QRS = 74^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quad\;PQRS = interior\;\;opposite\;\;\angle \\[3ex] Similarly: \\[3ex] \angle UQR = \angle PSR = 68^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quad\;PQRS = interior\;\;opposite\;\;\angle \\[3ex] But:\;\; \angle PSR = \angle PSQ + \angle QSR ...diagram \\[3ex] 68 = \angle PSQ + 40 \\[3ex] \angle PSQ + 40 = 68 \\[3ex] \angle PSQ = 68 - 40 \\[3ex] \angle PSQ = 28 \\[3ex] \angle PSQ = \angle PRQ = 28^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] But:\;\;\angle QRS = \angle PRQ + \angle PRS \\[3ex] 74 = 28 + \angle PRS \\[3ex] 28 + \angle PRS = 74 \\[3ex] \angle PRS = 74 - 28 \\[3ex] \angle PRS = 46^\circ $(95.) CSEC The diagram below, not drawn to scale, shows a circle, centre O The line DCE is a tangent to the circle. Angle ACE = 48° and angle OCB = 26° Calculate: (i) ∠ABC (ii) ∠AOC (iii) ∠BCD (iv) ∠BAC (v) ∠OAC (vi) ∠OAB$ (i) \\[3ex] \angle ACE = \angle ABC...\angle \;\;between\;\;tangent\;DCE\;\;and\;\;chord\;AC = \angle\;\;in\;\;the\;\;alternate\;\;segemnt \\[3ex] 48 = \angle ABC \\[3ex] \angle ABC = 48^\circ \\[3ex] (ii) \\[3ex] \angle AOC = 2 * \angle ABC ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOC = 2 * 48 \\[3ex] \angle AOC = 96^\circ \\[3ex] (iii) \\[3ex] \angle OCD = 90^\circ ...radius\;OC\;\perp\;tangent\;DCE\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] But:\;\;\angle OCD = \angle OCB + \angle BCD ...diagram \\[3ex] 90 = 26 + \angle BCD \\[3ex] 26 + \angle BCD = 90 \\[3ex] \angle BCD = 90 - 26 \\[3ex] \angle BCD = 64^\circ \\[3ex] (iv) \\[3ex] \angle BCD = \angle BAC...\angle \;\;between\;\;tangent\;DCE\;\;and\;\;chord\;BC = \angle\;\;in\;\;the\;\;alternate\;\;segemnt \\[3ex] 64 = \angle BAC \\[3ex] \angle BAC = 64^\circ \\[3ex] (v) \\[3ex] \underline{\triangle AOC} \\[3ex] \angle OAC = \angle OCA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOC \\[3ex] \angle OAC + \angle OCA + \angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] p + p + 96 = 180 \\[3ex] 2p = 180 - 96 \\[3ex] 2p = 84 \\[3ex] p = \dfrac{84}{2} \\[5ex] p = 42 \\[3ex] \therefore \angle OAC = \angle OCA = 42^\circ \\[3ex] (vi) \\[3ex] \angle BAC = \angle OAB + \angle OAC ...diagram \\[3ex] 64 = \angle OAB + 42 \\[3ex] \angle OAB + 42 = 64 \\[3ex] \angle OAB = 64 - 42 \\[3ex] \angle OAB = 22^\circ $(96.) ACT The circle shown below has diameter$\overline{AD}$, and points B and C lie on the circle. The measure of$\angle CAD$is$30^\circ$, and the measure of minor arc$\overset{\huge\frown}{CD}$is$60^\circ$What is the measure of minor arc$\overset{\huge\frown}{AC} F.\;\; 75^\circ \\[3ex] G.\;\; 90^\circ \\[3ex] H.\;\; 105^\circ \\[3ex] J.\;\; 120^\circ \\[3ex] K.\;\; 150^\circ \\[3ex]  \overline{AD}\;\;is\;\;a\;\;diameter \\[3ex] \angle ACD = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle ACD} \\[3ex] \angle ADC + \angle ACD + \angle CAD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] \angle ADC + 90 + 30 = 180 \\[3ex] \angle ADC + 120 = 180 \\[3ex] \angle ADC = 180 - 120 \\[3ex] \angle ADC = 60^\circ \\[3ex] \overset{\huge\frown}{AC} = 2 * \angle ADC ... intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] \overset{\huge\frown}{AC} = 2 * 60 \\[3ex] \overset{\huge\frown}{AC} = 120^\circ $(97.) WASSCE In the diagram,$\overline{RQ}$is a tangent to the circle,$|PS| = |SQ|$and$\angle SQR = 50^\circ$Calculate$\angle SRQ A.\:\: 30^\circ \\[3ex] B.\:\: 35^\circ \\[3ex] C.\:\: 40^\circ \\[3ex] D.\:\: 45^\circ \\[3ex]  \angle SRQ = \angle SPQ = 50^\circ ...\angle \:\:between\:\:tangent\;RQ\:\:and\:\:chord\;QS = \angle\:\:in\;\;the\:\:alternate\:\:segment \\[3ex] \angle SPQ = \angle SQP = 50^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ = \angle SPQ + \angle SQP ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle RSQ = 50 + 50 \\[3ex] \angle RSQ = 100^\circ \\[3ex] \angle SRQ + \angle RSQ + \angle SQR = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle SRQ \\[3ex] \angle SRQ + 100 + 50 = 180 \\[3ex] \angle SRQ + 150 = 180 \\[3ex] \angle SRQ = 180 - 150 \\[3ex] \angle SRQ = 30^\circ $(98.) GCSE A, B, C and D are points on the circumference of a circle, centre O. AC is a diameter of the circle. AC and BD intersect at E. Angle CAB = 25° Angle DEC = 100° Work out the size of angle DAC. You must show all your working.$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACB \\[3ex] 25 + 90 + \angle ACB = 180 \\[3ex] 115 + \angle ACB = 180 \\[3ex] \angle ACB = 180 - 115 \\[3ex] \angle ACB = 65^\circ \\[3ex] \angle ADB = \angle ACB = 65^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle AED + \angle DEC = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle AED + 100 = 180 \\[3ex] \angle AED = 180 - 100 \\[3ex] \angle AED = 80^\circ \\[3ex] \underline{\triangle AED} \\[3ex] \angle DAE + \angle AED + \angle ADE = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle ADE \\[3ex] \angle ADE = \angle ADB = 65^\circ...diagram \\[3ex] \implies \\[3ex] \angle DAE + 80 + 65 = 180 \\[3ex] \angle DAE + 145 = 180 \\[3ex] \angle DAE = 180 - 145 \\[3ex] \angle DAE = 35^\circ \\[3ex] \angle DAC = \angle DAE = 35^\circ ...diagram $(99.) NZQA In the diagram "alongside" (below): BC and BE are radii of the circle centre B GI is a tangent to the circle. Angle CEG is 38° Calculate the size, x, of angle BCE$ \angle BCE = \angle BEC = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BCE \\[3ex] \angle BEI = 90^\circ ...radius\;BE \perp tangent\;GEI\;\;at\;\;point\;\;of\;\;contact\;E \\[3ex] \angle CEG + \angle BEC + \angle BEI = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 38 + x + 90 = 180 \\[3ex] x + 128 = 180 \\[3ex] x = 180 - 128 \\[3ex] x = 52^\circ \\[3ex] $Common mistake:$\angle CEG = \angle EBC = 38^\circ...\angle \;\;between\;\;tangent\;\;and\;\;chord = \angle\;\;in\;\;alternate\;\;segment$This is not correct because$\angle EBC$is not in the alternate segment. The alternate segement theorem requires that the angle is formed by the intersection of two chords that touches the circumference of the circle (not the angle at the centre of the circle formed by the intersection of two radii) (100.) WASSCE In the diagram, O is the centre of the circle. If WX is parallel to YZ and$\angle WXY = 50^\circ$, find the value of: (i)$\angle WOZ$(ii)$\angle YEZ (i) \\[3ex] \underline{\triangle WXY} \\[3ex] \angle XWY + \angle WXY + \angle XYW = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle WXY \\[3ex] \angle XWY + 50 + 90 = 180 \\[3ex] \angle XWY + 140 = 180 \\[3ex] \angle XWY = 180 - 140 \\[3ex] \angle XWY = 40^\circ \\[3ex] \angle XWY = \angle WYZ = 40^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle WOZ = 2 * \angle WYZ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle WOZ = 2 * 40 \\[3ex] \angle WOZ = 80^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle OWE} \\[3ex] \angle WOE + \angle OWE + \angle OEW = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle OWE \\[3ex] \angle WOE = \angle WOZ = 80^\circ...diagram \\[3ex] \angle OWE = \angle XWY = 40^\circ \\[3ex] \implies \\[3ex] 80 + 40 + \angle OEW = 180 \\[3ex] 120 + \angle OEW = 180 \\[3ex] \angle OEW = 180 - 120 \\[3ex] \angle OEW = 60^\circ \\[3ex] \angle YEZ = \angle OEW = 60^\circ ...vertical\;\;\angle s\;\;are\;\;equal $(101.) GCSE P, M and S are points on a circle, centre O. RST is a tangent to the circle. Angle PSO = 48° MP = MS Work out the size of angle MST. Give reasons for each stage of your working. Construction: Draw the radius from the centre O to point P on the circumference$ \angle OPS = \angle OSP = 48^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POS \\[3ex] \angle OPS + \angle OSP + \angle POS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POS \\[3ex] 48 + 48 + \angle POS = 180 \\[3ex] 96 + \angle POS = 180 \\[3ex] \angle POS = 180 - 96 \\[3ex] \angle POS = 84^\circ \\[3ex] \angle POS = 2 * \angle PMS ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 84 = 2 * \angle PMS \\[3ex] 2 * \angle PMS = 84 \\[3ex] \angle PMS = \dfrac{84}{2} \\[5ex] \angle PMS = 42^\circ \\[3ex] \angle MST = \angle PMS = 42^\circ ... \angle \;\;between\;\;tangent\;RST\;\;and\;\;chord\;SM = \angle \;\;in\;\;the\;\;alternate\;\;segment $(102.) NZQA In the diagram below, the line MN passes through the centre of the circle, O Angle MQO is 71°, angle SNO is 37° and angle SRO is 75° (i) Find the size of angle p Justify your answer with clear geometric reasoning. (ii) Find the size of angle e Justify your answer with clear geometric reasoning.$ (i) \\[3ex] \angle OQM = \angle OMQ = 71^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OQM \\[3ex] \angle MQN = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle MNQ} \\[3ex] \angle QMN + \angle MQN + \angle MNQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNQ \\[3ex] \angle QMN = \angle OMQ = 71^\circ ...diagram \\[3ex] \angle MNQ = p \\[3ex] \implies \\[3ex] 71 + 90 + p = 180 \\[3ex] 161 + p = 180 \\[3ex] p = 180 - 161 \\[3ex] p = 19^\circ \\[3ex] $Construction: Join the radius from centre O to point S on the circumference$ (ii) \\[3ex] \underline{Same\;\;Radius} \\[3ex] |OR| = |OM| = |OQ| = |ON| = |OS| \\[3ex] \implies \\[3ex] \underline{Base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle s} \\[3ex] \triangle ORM:\;\;\;\angle ORM = \angle OMR = e \\[3ex] \triangle OMQ:\;\;\;\angle OMQ = \angle OQM = 71^\circ \\[3ex] \triangle OQN:\;\;\;\angle OQN = \angle ONQ = p \\[3ex] \triangle ONS:\;\;\;\angle ONS = \angle OSN = 37^\circ \\[3ex] \triangle OSR:\;\;\;\angle OSR = \angle ORS = 75^\circ \\[3ex] \underline{Sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle = 180^\circ} \\[3ex] \triangle ORM:\;\;\;e + e + \angle ROM = 180 \\[3ex] 2e + \angle ROM = 180 \\[3ex] \angle ROM = 180 - 2e \\[3ex] \triangle OMQ:\;\;\; 71 + 71 + \angle MOQ = 180 \\[3ex] 142 + \angle MOQ = 180 \\[3ex] \angle MOQ = 180 - 142 \\[3ex] \angle MOQ = 38^\circ \\[3ex] \triangle OQN:\;\;\; p + p + \angle QON = 180 \\[3ex] 2p + \angle QON = 180 \\[3ex] 2(19) + \angle QON = 180 \\[3ex] 38 + \angle QON = 180 \\[3ex] \angle QON = 180 - 38 \\[3ex] \angle QON = 142^\circ \\[3ex] \triangle NOS:\;\;\; 37 + 37 + \angle NOS = 180 \\[3ex] 74 + \angle NOS = 180 \\[3ex] \angle NOS = 180 - 74 \\[3ex] \angle NOS = 106^\circ \\[3ex] \triangle SOR:\;\;\;75 + 75 + \angle SOR = 180 \\[3ex] 150 + \angle SOR = 180 \\[3ex] \angle SOR = 180 - 150 \\[3ex] \angle SOR = 30^\circ \\[3ex] \underline{\angle s\;\;at\;\;a\;\;point = 360^\circ} \\[3ex] \angle ROM + \angle MOQ + \angle QON + \angle NOS + \angle SOR = 360 \\[3ex] (180 - 2e) + 38 + 142 + 106 + 30 = 360 \\[3ex] 180 - 2e + 316 = 360 \\[3ex] 496 - 2e = 360 \\[3ex] 496 - 360 = 2e \\[3ex] 136 = 2e \\[3ex] 2e = 136 \\[3ex] e = \dfrac{136}{2} \\[5ex] e = 68^\circ $(103.) WASSCE In the diagram,$\angle RTS = 28^\circ$,$\angle VRM = 46^\circ$, MQ is a tangent to the circle VRSTU at the point R. Find$\angle VUS \angle SRQ = 28^\circ ...\angle \;\;between\;\;tangent\;MRQ\;\;and\;\;chord\;RS = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle VRM + \angle VRS + \angle SRQ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 46 + \angle VRS + 28 = 180 \\[3ex] 74 + \angle VRS = 180 \\[3ex] \angle VRS = 180 - 74 \\[3ex] \angle VRS = 106^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;VUSR} \\[3ex] \angle U + \angle R = 180^\circ ...opposite\;\;\angle s\;\;of\;\;acyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle U = \angle VUS = x \\[3ex] \angle R = \angle VRS = 106^\circ \\[3ex] \implies \\[3ex] x + 106 = 180 \\[3ex] x = 180 - 106 \\[3ex] x = 74^\circ $(104.) NZQA In the diagram below, angle SRO is 75° Find an expression for z in terms of y Justify your answer with clear geometric reasoning. Construction: Join the radius from the centre O to point S$ \underline{Same\;\;Radii} \\[3ex] |OR| = |OS| = |ON| \\[3ex] \implies \\[3ex] \underline{\triangle SOR} \\[3ex] \angle ORS = \angle OSR = 75^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SOR \\[3ex] \angle ORS + \angle OSR + \angle SOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOR \\[3ex] 75 + 75 + \angle SOR = 180 \\[3ex] 150 + \angle SOR = 180 \\[3ex] \angle SOR = 180 - 150 \\[3ex] \angle SOR = 30^\circ \\[3ex] \angle RSN = \angle OSR + \angle OSN = z ...diagram \\[3ex] 75 + \angle OSN = z \\[3ex] \angle OSN = z - 75 \\[3ex] \underline{\triangle SON} \\[3ex] \angle OSN = \angle ONS = z - 75 ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SON \\[3ex] \angle SOM = \angle OSN + \angle ONS \\[3ex] ...exterior\;\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s\\[3ex] \angle SOM = \angle SOR + \angle ROM \\[3ex] \angle ROM = y \\[3ex] \therefore \angle SOM = 30 + y \\[3ex] \implies \\[3ex] 30 + y = (z - 75) + (z - 75) \\[3ex] 30 + y = z - 75 + z - 75 \\[3ex] 30 + y = 2z - 150 \\[3ex] 2z - 150 = 30 + y \\[3ex] 2z = 30 + y + 150 \\[3ex] 2z = y + 180 \\[3ex] z = \dfrac{y + 180}{2} \\[5ex] OR \\[3ex] \angle OSN + \angle ONS + \angle SON = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SON \\[3ex] (z - 75) + (z - 75) + \angle SON = 180 \\[3ex] z - 75 + z - 75 + \angle SON = 180 \\[3ex] 2z - 150 + \angle SON = 180 \\[3ex] \angle SON = 180 - 2z + 150 \\[3ex] \angle SON = 330 - 2z \\[3ex] \angle ROM + \angle SOR + \angle SON = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] y + 30 + (330 - 2z) = 180 \\[3ex] y + 30 + 330 - 2z = 180 \\[3ex] y + 360 - 180 = 2z \\[3ex] y + 180 = 2z \\[3ex] 2z = y + 180 \\[3ex] z = \dfrac{y + 180}{2} $(105.) CSEC The diagram below, not drawn to scale, shows a circle, centre O The line BC is a tangent to the circle at B Angle CBD = 42° and angle OBE = 20° Calculate, giving a reason for EACH step of your answer, the measure of: (i) ∠BOE (ii) ∠OED (iii) ∠BFE$ (i) \\[3ex] \underline{\triangle BOE} \\[3ex] \angle BEO = \angle EBO = 20^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOE \\[3ex] \angle BEO + \angle EBO + \angle BOE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOE \\[3ex] 20 + 20 + \angle BOE = 180 \\[3ex] 40 + \angle BOE = 180 \\[3ex] \angle BOE = 180 - 40 \\[3ex] \angle BOE = 140^\circ \\[3ex] (ii) \\[3ex] 42^\circ = \angle BED ...\angle\;\;between\;\;tangent\;BC\;\;and\;\;chord\;BD = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle BED = 42^\circ \\[3ex] Also: \\[3ex] \angle BED = \angle BEO + \angle OED ...diagram \\[3ex] 42 = 20 + \angle OED \\[3ex] 20 + \angle OED = 42 \\[3ex] \angle OED = 42 - 20 \\[3ex] \angle OED = 22^\circ \\[3ex] (iii) \\[3ex] Reflex\;\;\angle BOE + Obtuse\;\;\angle BOE = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle BOE + 140 = 360 \\[3ex] Reflex\;\;\angle BOE = 360 - 140 \\[3ex] Reflex\;\;\angle BOE = 220^\circ \\[3ex] Reflex\;\;\angle BOE = 2 * \angle BFE \\[3ex] 220 = 2 * \angle BFE \\[3ex] 2 * \angle BFE = 220 \\[3ex] \angle BFE = \dfrac{220}{2} \\[5ex] \angle BFE = 110^\circ $(106.) GCSE A, B, C and D are points on the circumference of a circle, centre O Angle AOC = y Find the size of angle ABC in terms of y Give a reason for each stage of your working.$ Reflex\;\;\angle AOC + Obtuse\;\;\angle AOC = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle AOC + y = 360 \\[3ex] Reflex\;\;\angle AOC = 360 - y \\[3ex] Also: \\[3ex] Reflex\;\;\angle AOC = 2 * \angle ABC \\[3ex] ...\angle\;\;at\;\;center = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 360 - y = 2 * \angle ABC \\[3ex] 2 * \angle ABC = 360 - y \\[3ex] \angle ABC = \dfrac{360 - y}{2} $(107.) NZQA A, B, and C are on the circumference of a circle with centre O BOC is a diameter. QCP is a tangent to the circle. Angle ACP = 40° (i) Find the size of angle ACO Justify your answer wuth clear geometric reasoning. (ii) Find the size of angle OAB Justify your answer wuth clear geometric reasoning.$ (i) \\[3ex] \angle OCP = 90^\circ... \\[3ex] radius\;OC \perp tangent\;PCQ \;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] But: \\[3ex] \angle OCP = \angle ACO + \angle ACP ...diagram \\[3ex] 90 = \angle ACO + 40 \\[3ex] \angle ACO + 40 = 90 \\[3ex] \angle ACO = 90 - 40 \\[3ex] \angle ACO = 50^\circ \\[3ex] (ii) \\[3ex] \angle ACP = \angle ABC = 40^\circ ... \angle\;\;between\;\;tangent\;PCQ\;\;and\;\;chord\;AC = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle ABC = \angle OBA = 40^\circ ...diagram \\[3ex] Because\;\;|OA| = |OB|...same\;\;radius: \\[3ex] \angle OAB = \angle OBA = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB $(108.) CSEC The diagram below, not drawn to scale, shows a circle, centre O The line UVW is a tangent to the circle, ZOXW is a straight line and angle UOX = 70° (i) Calculate, showing working where necessary, the measure of angle a) OUZ b) UVY c) UWO (ii) Name the triangle in the diagram which is congruent to triangle a) ZOU b) YXU$ (i) \\[3ex] a) \\[3ex] \angle UOX = \angle OZU + \angle OUZ \\[3ex] ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle OZU = \angle OUZ = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ZOU \\[3ex] \implies \\[3ex] 70 = p + p \\[3ex] 70 = 2p \\[3ex] 2p = 70 \\[3ex] p = \dfrac{70}{2} \\[5ex] p = 35 \\[3ex] \therefore \angle OZU = \angle OUZ = 35^\circ \\[3ex] b) \\[3ex] \angle OZU = \angle OYX = 35^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle OUW = 90^\circ \\[3ex] ...radius\;OU \perp tangent\;UVW\;\;at\;\;point\;\;of\;\;contact\;U \\[3ex] \underline{\triangle UYV} \\[3ex] \angle UYV = \angle OYX = 35^\circ ...diagram \\[3ex] \angle YUV = \angle OUW = 90^\circ...diagram \\[3ex] \angle UYV + \angle YUV + \angle UVY = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle UYV \\[3ex] \implies \\[3ex] 35 + 90 + \angle UVY = 180 \\[3ex] 125 + \angle UVY = 180 \\[3ex] \angle UVY = 180 - 125 \\[3ex] \angle UVY = 55^\circ \\[3ex] c) \\[3ex] \underline{\triangle UOW} \\[3ex] \angle UOW = \angle UOX = 70^\circ ...diagram \\[3ex] \angle UOW + \angle OUW + \angle UWO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle UOW \\[3ex] \implies \\[3ex] 70 + 90 + \angle UWO = 180 \\[3ex] 160 + \angle UWO = 180 \\[3ex] \angle UWO = 180 - 160 \\[3ex] \angle UWO = 20^\circ \\[3ex] (ii) \\[3ex] a) \\[3ex] \underline{\triangle ZOU} \\[3ex] |OZ| = |OU|...same\;\;radius \\[3ex] \angle ZOU + \angle UOX = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ZOU + 70 = 180 \\[3ex] \angle ZOU = 180 - 70 \\[3ex] \angle ZOU = 110^\circ \\[3ex] \underline{\triangle YOX} \\[3ex] |OY| = |OX|...same\;\;radius \\[3ex] \angle YOX + \angle UOX = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle YOX + 70 = 180 \\[3ex] \angle YOX = 180 - 70 \\[3ex] \angle YOX = 110^\circ \\[3ex] Because: \\[3ex] |ZO| = |YO| \;\;and \\[3ex] \angle ZOU = \angle YOX \;\;and \\[3ex] |OU| = |OX| \\[3ex] This\;\;implies\;\;that \\[3ex] \triangle YOX \cong \triangle ZOU ...Side-Angle-Side \\[3ex] \underline{\triangle YXU} \\[3ex] \angle UYX = \angle UYV = 35^\circ...diagram \\[3ex] \angle YXU = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] |YU| = diameter \\[3ex] \underline{\triangle ZUX} \\[3ex] \angle UZX = \angle UYX = 35^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ZUX = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] |ZX| = diameter \\[3ex] Because: \\[3ex] \angle UYX = \angle UZX \;\;and \\[3ex] \angle ZUX = \angle YXU \;\;and \\[3ex] |YU| = |ZX| \\[3ex] This\;\;implies\;\;that \\[3ex] \triangle ZUX \cong \triangle YXU ...Angle-Angle-Side $(109.) CSEC In the diagram below, not drawn to scale, PQ is a tangent to the circle PTSR, so that ∠RPQ = 46° RQP = 32° and TRQ is a straight line. Calculate, giving a reason for EACH step of your answer, (i) ∠PTR (ii) ∠TPR (iii) ∠TSR$ (i) \\[3ex] 46^\circ = \angle PTR ...\angle\;\;between\;\;tangent\;PQ\;\;and\;\;chord\;PR = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle PTR = 46^\circ \\[3ex] (ii) \\[3ex] \angle TRP = \angle RPQ + \angle RQP ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle TRP = 46 + 32 \\[3ex] \angle TRP = 78^\circ \\[3ex] \underline{\triangle TPR} \\[3ex] \angle PTR + \angle TPR + \angle TRP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TPR \\[3ex] 46 + \angle TPR + 78 = 180 \\[3ex] 124 + \angle TPR = 180 \\[3ex] \angle TPR = 180 - 124 \\[3ex] \angle TPR = 56^\circ \\[3ex] (iii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;TPRS} \\[3ex] \angle P + \angle S = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle P = \angle TPR = 56^\circ...diagram \\[3ex] \angle S = \angle TSR ...diagram \\[3ex] \implies \\[3ex] 56 + \angle TSR = 180 \\[3ex] \angle TSR = 180 - 56 \\[3ex] \angle TSR = 124^\circ $(110.) WASSCE In the diagram, O is the centre of the circle,$\angle XOZ = (10m)^\circ$and$\angle XWZ = m^\circ$Calculate the value of m$ A.\;\; 30 \\[3ex] B.\;\; 36 \\[3ex] C.\;\; 40 \\[3ex] D.\;\; 72 \\[3ex]  Reflex\;\;\angle XOZ + Obtuse\;\;\angle O = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle XOZ + 10m = 360 \\[3ex] Reflex\;\;\angle XOZ = 360 - 10m ...eqn.(1) \\[3ex] Also: \\[3ex] Reflex\;\;\angle XOZ = 2 * \angle XWZ \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle XOZ = 2 * m ...eqn.(2) \\[3ex] eqn.(1) = eqn.(2) \implies \\[3ex] 360 - 10m = 2 * m \\[3ex] 2 * m = 360 - 10m \\[3ex] 2m + 10m = 360 \\[3ex] 12m = 360 \\[3ex] m = \dfrac{360}{12} \\[5ex] m = 30^\circ $(111.) NSC In the diagram, PQRS is a cyclic quadrilateral. ST is a tangent to the circle at S and chord SR is produced to V PQ = QR,$\hat{S_1} = 42^\circ$and$\hat{S_2} = 108^\circ$Determine, with reasons, the size of the following angles: (111.1)$\hat{Q}$(111.2)$\hat{R_2}$(111.3)$\hat{P_2}$(111.4)$\hat{R_3} (111.1) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \hat{Q} + \hat{S_2} = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \hat{Q} + 108 = 180 \\[3ex] \hat{Q} = 180 - 108 \\[3ex] \hat{Q} = 72^\circ \\[3ex] (111.2) \\[3ex] \underline{\triangle PQR} \\[3ex] \hat{P_1} = \hat{R_2} = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PQR \\[3ex] \hat{P_1} + \hat{Q} + \hat{R_2} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] k + 72 + k = 180 \\[3ex] 2k + 72 = 180 \\[3ex] 2k = 180 - 72 \\[3ex] 2k = 108 \\[3ex] k = \dfrac{108}{2} \\[5ex] k = 54 \\[3ex] \therefore \hat{P_1} = \hat{R_2} = 54^\circ \\[3ex] (111.3) \\[3ex] 42^\circ = \hat{P_2} ...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SR = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \hat{P_2} = 42^\circ \\[3ex] (111.4) \\[3ex] \hat{R_3} = \hat{P}...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] \hat{P} = \hat{P_1} + \hat{P_2} ...diagram \\[3ex] \implies \\[3ex] \hat{R_3} = \hat{P_1} + \hat{P_2} \\[3ex] \hat{R_3} = 54 + 42 \\[3ex] \hat{R_3} = 96^\circ $(112.) ACT From point A outside a circle and in the same plane as the circle, 2 rays are drawn tangent to the circle with the points of tangency labeled B and C, respectively. Segment$\overline{BC}$is then drawn to form$\triangle ABC$If$\angle A$measures$70^\circ$, what is the measure of$\angle ABC$?$ F.\;\; 70^\circ \\[3ex] G.\;\; 55^\circ \\[3ex] H.\;\; 40^\circ \\[3ex] J.\;\; 35^\circ \\[3ex] $K. Cannot be determined from the given information Let us draw the diagram for this question AB = AC ...the two tangents: AB and AC, drawn from the same external point:A are equal in length$ \implies \triangle ABC\;\;is\;\;an\;\;isosceles\;\;triangle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ABC + \angle ACB + \angle BAC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] Let\;\; \angle ABC = \angle ACB = p \\[3ex] \angle BAC = \angle A = 70^\circ \\[3ex] \implies \\[3ex] p + p + 70 = 180 \\[3ex] 2p = 180 - 70 \\[3ex] 2p = 110 \\[3ex] p = \dfrac{110}{2} \\[5ex] p = 55 \\[3ex] \therefore \angle ABC = 55^\circ $(113.) CSEC The diagram below, not drawn to scale, shows a circle with centre O EBC is a tangent to the circle. OBA = 40° and ∠OBF = 35° Calculate, giving reasons for your answer, the measure of (i) ∠EBF (ii) ∠BOA (iii) ∠AFB (iv) ∠OAF$ (i) \\[3ex] \angle EBF + \angle OBF = 90^\circ ...radius\;OB \perp tangent\;EBC\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \angle EBF + 35 = 90 \\[3ex] \angle EBF = 90 - 35 \\[3ex] \angle EBF = 55^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle BOA} \\[3ex] \angle OBA = \angle OAB = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOA \\[3ex] \angle OBA + \angle OAB + \angle BOA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOA \\[3ex] 40 + 40 + \angle BOA = 180 \\[3ex] 80 + \angle BOA = 180 \\[3ex] \angle BOA = 180 - 80 \\[3ex] \angle BOA = 100^\circ \\[3ex] (iii) \\[3ex] \angle BOA = 2 * \angle AFB ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 100 = 2 * \angle AFB \\[3ex] 2 * \angle AFB = 100 \\[3ex] \angle AFB = \dfrac{100}{2} \\[5ex] \angle AFB = 50^\circ \\[3ex] (iv) \\[3ex] \underline{\triangle BFA} \\[3ex] \angle ABF + \angle AFB + \angle BAF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BFA \\[3ex] \angle ABF = 35 + 40 ...diagram \\[3ex] \angle ABF = 75^\circ \\[3ex] \angle BAF = \angle OAB + \angle OAF ...diagram \\[3ex] \angle BAF = 40 + \angle OAF \\[3ex] \implies \\[3ex] 75 + 50 + (40 + \angle OAF) = 180 \\[3ex] 125 + 40 + \angle OAF = 180 \\[3ex] 165 + \angle OAF = 180 \\[3ex] \angle OAF = 180 - 165 \\[3ex] \angle OAF = 15^\circ $(114.) WASSCE The diagram shows a circle centre O If$\angle ZYW = 33^\circ$, find$\angle ZWX A.\;\; 33^\circ \\[3ex] B.\;\; 57^\circ \\[3ex] C.\;\; 90^\circ \\[3ex] D.\;\; 100^\circ \\[3ex]  \angle ZXW = \angle ZYW = 33^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle WZX = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle WZX} \\[3ex] \angle ZWX + \angle WZX + \angle ZXW = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WZX \\[3ex] \angle ZWX + 90 + 33 = 180 \\[3ex] \angle ZWX + 123 = 180 \\[3ex] \angle ZWX = 180 - 123 \\[3ex] \angle ZWX = 57^\circ $(115.) WASSCE In the diagram,$\overline{TS}$is a tangent to the circle at S If O is the centre of the circle,$\angle TSP = 21^\circ$and$\angle RQP = 100^\circ$, find with reasons: (i)$\angle SPR$(ii)$\angle QSR (i) \\[3ex] 21^\circ = \angle SQP...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SP = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle SQP = 21^\circ \\[3ex] \angle SRP = \angle SQP = 21^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle S + \angle Q = 180^\circ...opposite\;\;\angle s\;\;of\;\;Cyclic\;\;Quad\;\;RQPS\;\;are\;\;supplementary \\[3ex] \angle S = \angle RSP ...diagram \\[3ex] \angle Q = \angle RQP = 100^\circ...diagram \\[3ex] \implies \\[3ex] \angle RSP + 100 = 180 \\[3ex] \angle RSP = 180 - 100 \\[3ex] \angle RSP = 80^\circ \\[3ex] \underline{\triangle SRP} \\[3ex] \angle RSP + \angle SRP + \angle SPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRP \\[3ex] 80 + 21 + \angle SPR = 180 \\[3ex] 101 + \angle SPR = 180 \\[3ex] \angle SPR = 180 - 101 \\[3ex] \angle SPR = 79^\circ \\[3ex] (ii) \\[3ex] \angle SQR = \angle SPR = 79^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle SRQ = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle QSR + \angle SQR + \angle SRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] \angle QSR + 79 + 90 = 180 \\[3ex] \angle QSR + 169 = 180 \\[3ex] \angle QSR = 180 - 169 \\[3ex] \angle QSR = 11^\circ $(116.) GCSE The points B, C, Y and X lie on a circle. AXY and ABC are straight lines. AX = 12cm XY = xcm AB = xcm BC = 4cm (a) Show that$x^2 - 8x - 144 = 0$(b) Find the length of AC. Show your working clearly. Give your answer correct to 3 significant figures.$ (a) \\[3ex] 12(12 + x) = x(x + 4) ...Intersecting\;\;Secants\;\;Theorem \\[3ex] 144 + 12x = x^2 + 4x \\[3ex] x^2 + 4x = 144 + 12x \\[3ex] x^2 + 4x - 12x - 144 = 0 \\[3ex] x^2 - 8x - 144 = 0 \\[3ex] (b) \\[3ex] x^2 - 8x - 144 = 0 \\[3ex] Compare\;\;to\;\;general\;\;form:\;\;ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -8 \\[3ex] c = -144 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} ...Quadratic\;\;Formula \\[5ex] x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-144)}}{2(1)} \\[5ex] x = \dfrac{8 \pm \sqrt{64 + 576}}{2} \\[5ex] x = \dfrac{8 \pm \sqrt{640}}{2} \\[5ex] x = \dfrac{8 \pm \sqrt{64} * \sqrt{10}}{2} \\[5ex] x = \dfrac{8 \pm 8 * \sqrt{10}}{2} \\[5ex] x = \dfrac{8(1 \pm \sqrt{10}}{2} \\[5ex] x = 4(1 \pm \sqrt{10}) \\[3ex] x = 4(1 \pm 3.16227766) \\[3ex] x = 4(1 + 3.16227766) \;\;\;OR\;\;\; x = 4(1 - 3.16227766) \\[3ex] x = 4(4.16227766) \;\;\;OR\;\;\; x = 4(-2.16227766) \\[3ex] $Because the length cannot be negative, discard the negative value$ x = 16.64911064 \\[3ex] |AC| = |AB| + |BC| ... diagram \\[3ex] = x + 4 \\[3ex] = 16.64911064 + 4 \\[3ex] = 20.64911064 \\[3ex] \approx 20.6\;cm ...to\;\;3\;\;significant\;\;figures $(117.) WASSCE In the diagram, RP is a diameter of the circle RSP RP is produced to T and TS is a tangent to the circle at S If ∠PRS = 24°, claculate the value of ∠STR$ A.\;\; 24^\circ \\[3ex] B.\;\; 42^\circ \\[3ex] C.\;\; 48^\circ \\[3ex] D.\;\; 66^\circ \\[3ex]  \angle RSP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PST = 24^\circ ...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SP = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle RST = \angle RSP + \angle PST ...diagram \\[3ex] \angle RST = 90 + 24 \\[3ex] \angle RST = 114^\circ \\[3ex] \underline{\triangle RST} \\[3ex] \angle TRS + \angle RST + \angle STR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RST \\[3ex] \angle TRS = \angle PRS = 24^\circ ...diagram \\[3ex] \implies \\[3ex] 24 + 114 + \angle STR = 180 \\[3ex] 138 + \angle STR = 180 \\[3ex] \angle STR = 180 - 138 \\[3ex] \angle STR = 42^\circ $(118.) CSEC P, Q, R and S are four points on the circumference of the circle shown below. Angle QRS = 58° Using the geometrical properties of a circle to give reasons for each step of your answer, determine the measure of (i) ∠SPQ (ii) ∠OQS$ (i) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;SRQP} \\[3ex] \angle R + \angle P = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle R = \angle QRS = 58^\circ \\[3ex] \angle P = \angle SPQ ...diagram \\[3ex] \implies \\[3ex] 58 + \angle SPQ = 180 \\[3ex] \angle SPQ = 180 - 58 \\[3ex] \angle SPQ = 122^\circ \\[3ex] OR \\[3ex] Obtuse\;\;\angle SOQ = 2 * \angle SRQ \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle SOQ = 2 * 58 \\[3ex] Obtuse\;\;\angle SOQ = 116^\circ \\[3ex] Reflex\;\;\angle SOQ + Obtuse\;\;\angle SOQ = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle SOQ + 116 = 360 \\[3ex] Reflex\;\;\angle SOQ = 360 - 116 \\[3ex] Reflex\;\;\angle SOQ = 244^\circ \\[3ex] Also: \\[3ex] Reflex\;\;\angle SOQ = 2 * \angle SPQ \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 244 = 2 * \angle SPQ \\[3ex] 2 * \angle SPQ = 244 \\[3ex] \angle SPQ = \dfrac{244}{2} \\[5ex] \angle SPQ = 122^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle SOQ} \\[3ex] \angle OSQ = \angle OQS = k ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SOQ \\[3ex] \angle OSQ + \angle OQS + Obtuse\;\;\angle SOQ = 180^\circ \\[3ex] ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOQ \\[3ex] k + k + 116 = 180 \\[3ex] 2k = 180 - 116 \\[3ex] 2k = 64 \\[3ex] k = \dfrac{64}{2} \\[5ex] k = 32 \\[3ex] \therefore \angle OSQ = \angle OQS = 32^\circ $(119.) GCSE The diagram shows a circle, centre O, with points B and D on the circumference The line AC touches the circle at B OB is parallel to DC and angle OAB = 49° (i) Write down the mathematical name of the line OB (ii) Write down the reason why angle ABO is 90° (iii) Find angle AOB (iv) Write down the reason why angle ADC = angle AOB (v) Complete the statement using a mathematical word. Triangle AOB is ..................................... to triangle ADC (vi) AB = 5.4 cm Calculate (a) OB (b) OA (c) the area of triangle AOB$ (i) \\[3ex] Line\;\;OB\;\;is\;\;the\;\;radius\;\;of\;\;the\;\;circle \\[3ex] (ii) \\[3ex] \angle ABO = 90^\circ ...radius\;OB \perp tangent\;ABC\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] (iii) \\[3ex] \angle AOB + \angle OAB + \angle ABO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB + 49 + 90 = 180 \\[3ex] \angle AOB + 139 = 180 \\[3ex] \angle AOB = 180 - 139 \\[3ex] \angle AOB = 41^\circ \\[3ex] (iv) \\[3ex] \angle ADC = \angle AOB = 41^\circ ... corresponding\;\;\angle s\;\;between\;\;parallel\;\;lines\;OB\;\;and\;\;DC,\;\;are\;\;equal \\[3ex] (v) \\[3ex] $Triangle AOB is similar to triangle ADC The remaining questions are Trigonometry (Triangles)$ (vi) \\[3ex] (a) \\[3ex] SOHCAHTOA \\[3ex] opp = |OB| \\[3ex] adj = |AB| \\[3ex] hyp = |OA| \\[3ex] \tan 49 = \dfrac{|OB|}{5.4} \\[5ex] 5.4 * \tan 49 = |OB| \\[3ex] |OB| = 5.4 * \tan 49 \\[3ex] |OB| = 5.4(1.150368407) \\[3ex] |OB| = 6.211989399\;cm \\[3ex] (b) \\[3ex] \cos 49 = \dfrac{5.4}{|OA|} \\[5ex] |OA| * \cos 49 = 5.4 \\[3ex] |OA| = \dfrac{5.4}{\cos 49} \\[5ex] |OA| = \dfrac{5.4}{0.656059029} \\[5ex] |OA| = 8.230966668\;cm \\[3ex] OR \\[3ex] hyp^2 = opp^2 + adj^2 ...Pythagorean\;\;Theorem \\[3ex] |OA|^2 = |OB|^2 + |AB|^2 \\[3ex] |OA|^2 = 6.211989399^2 + 5.4^2 \\[3ex] |OA|^2 = 38.58881229 + 29.16 \\[3ex] |OA|^2 = 67.74881229 \\[3ex] |OA| = \sqrt{67.74881229} \\[3ex] |OA| = 8.230966668\;cm \\[3ex] (c) \\[3ex] Area\;\;of\;\;\triangle AOB \\[3ex] = \dfrac{1}{2} * |OA| * |AB| * \sin \angle OAB \\[5ex] = \dfrac{1}{2} * 8.230966668 * 5.4 * \sin 49 \\[5ex] = 0.5 * 8.230966668 * 5.4 * 0.7547095802 \\[3ex] = 16.77237138\;cm^2 \\[3ex] OR \\[3ex] Area\;\;of\;\;\triangle AOB \\[3ex] = \dfrac{1}{2} * |AB| * |OB| \\[5ex] = 0.5 * 5.4 * 6.211989399 \\[3ex] = 16.77237138\;cm^2 $(120.) NZQA Amy draws another pentagon which is cyclic but not regular. PT is parallel to EN PXT is a diameter X is the centre of the circle (i) Amy thinks that angle PAT is 90° Is she correct? Give a geometric reason. (ii) Calculate the size of angle NXE Give geometric reasons. (iii) Suppose angle XPE = w It can be shown that angle NXE = 4w = 180° From this formula, what does this tell you about the size of angle w?$ (i) \\[3ex] Yes,\;\;she\;\;is\;\;correct \\[3ex] \angle PAT = 90^\circ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] $Construction: Draw the radii: from point X on the centre to point E on the circumference; and from point X on the centre to point N on the circumference of the circle$ (ii) \\[3ex] \angle XEP = \angle XPE = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PXE \\[3ex] \angle XEP + \angle XPE + \angle PXE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PXE \\[3ex] 52 + 52 + \angle PXE = 180 \\[3ex] 104 + \angle PXE = 180 \\[3ex] \angle PXE = 180 - 104 \\[3ex] \angle PXE = 76^\circ \\[3ex] \angle XEN = \angle PXE = 76^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle XNE = \angle XEN = 76^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XEN \\[3ex] \angle XNE + \angle XEN + \angle NXE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XEN \\[3ex] 76 + 76 + \angle NXE = 180 \\[3ex] 152 + \angle NXE = 180 \\[3ex] \angle NXE = 180 - 152 \\[3ex] \angle NXE = 28^\circ \\[3ex] (iii) \\[3ex] \angle XPE = w \\[3ex] \angle NXE = 4w - 180 \\[3ex] \angle NXE \gt 0 \\[3ex] \implies \\[3ex] 4w - 180 \gt 0 \\[3ex] 4w \gt 0 + 180 \\[3ex] 4w \gt 180 \\[3ex] w \gt \dfrac{180}{4} \\[5ex] w \gt 45^\circ $(121.) WASSCE In the diagram above, PTR is a tangent to the circle with centre O If ∠TON = 108°, calculate the size of ∠PTN$ A.\;\; 132^\circ \\[3ex] B.\;\; 126^\circ \\[3ex] C.\;\; 108^\circ \\[3ex] D.\;\; 102^\circ \\[3ex]  \angle OTN = \angle ONT = k...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle TON \\[3ex] \angle OTN + \angle ONT + \angle TON = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TON \\[3ex] k + k + 108 = 180 \\[3ex] 2k = 180 - 108 \\[3ex] 2k = 72 \\[3ex] k = \dfrac{72}{2} \\[5ex] k = 36 \\[3ex] \therefore \angle OTN = \angle ONT = 36^\circ \\[3ex] \angle OTP = 90^\circ...radius\;OT \perp tangent\;PTR\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle PTN = \angle OTP + \angle OTN ...diagram \\[3ex] \angle PTN = 90 + 36 \\[3ex] \angle PTN = 126^\circ $(122.) GCSE (a) A, B, C and D are points on the circle. AD is parallel to BC The chords AC and BD intersect at X Find the value of u and the value of v (b) F, G and H are points on the circle, centre O Find the value of p$ (a) \\[3ex] u = 35^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle DBC = \angle ADB = u \\[3ex] ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;AD\;\;and\;\;BC\;\;are\;\;equal \\[3ex] \angle DAX + \angle AXD + \angle ADX = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AXD \\[3ex] \angle ADX = \angle ADB = u = 35^\circ \\[3ex] \implies \\[3ex] 35 + v + 35 = 180 \\[3ex] v + 70 = 180 \\[3ex] v = 180 - 70 \\[3ex] v = 110^\circ \\[3ex] (b) \\[3ex] Reflex\;\;\angle FOG + Obtuse\;\;\angle FOG = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] 210 + Obtuse\;\;\angle FOG = 360 \\[3ex] Obtuse\;\;\angle FOG = 360 - 210 \\[3ex] Obtuse\;\;\angle FOG = 150^\circ \\[3ex] Also: \\[3ex] Obtuse\;\;\angle FOG = 2 * \angle FHG \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle FHP = p ...diagram \\[3ex] 150 = 2 * p \\[3ex] \implies \\[3ex] 2p = 150 \\[3ex] p = \dfrac{150}{2} \\[5ex] p = 75^\circ $(123.) CSEC In the diagram below, not drawn to scale, LM is a tangent to the circle at the point, T O is the centre of the circle and angle ∠MTS = 23° Calculate the size of each of the following angles, giving reasons for your answer. a) angle TPQ b) angle MTQ c) angle TQS d) angle SRQ$ a) \\[3ex] \angle TPQ = 90\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] b) \\[3ex] \angle MTQ = 90^\circ ...radius\;OT \perp tangent\;LTM\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex]  c) \\[3ex] 23^\circ = \angle TQS ...\angle\;\;between\;\;tangent\;LTM\;\;and\;\;chord\;TS = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle TQS = 23^\circ \\[3ex] d) \\[3ex] \angle MTS + \angle STQ = \angle MTQ ...diagram \\[3ex] 23 + \angle STQ = 90 \\[3ex] \angle STQ = 90 - 23 \\[3ex] \angle STQ = 67^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;TQRS} \\[3ex] \angle T + \angle R = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle T = \angle STQ = 67^\circ ...diagram \\[3ex] \angle R = \angle SRQ ...diagram \\[3ex] \implies \\[3ex] 67 + \angle SRQ = 180 \\[3ex] \angle SRQ = 180 - 67 \\[3ex] \angle SRQ = 113^\circ $(124.) WASSCE In the diagram, O is the centre of the circle, ∠OQR = 32° and ∠TPQ = 15° Calculate: (i) ∠QPR (ii) ∠TQO$ (i) \\[3ex] \angle OQR = \angle ORQ = 32^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle QOR \\[3ex] \angle OQR + \angle ORQ + \angle QOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QOR \\[3ex] 32 + 32 + \angle QOR = 180 \\[3ex] 64 + \angle QOR = 180 \\[3ex] \angle QOR = 180 - 64 \\[3ex] \angle QOR = 116^\circ \\[3ex] \angle QOR = 2 * \angle QPR ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 116 = 2 * \angle QPR \\[3ex] 2 * \angle QPR = 116 \\[3ex] \angle QPR = \dfrac{116}{2} \\[5ex] \angle QPR = 58^\circ \\[3ex] (ii) \\[3ex]  \underline{Cyclic\;\;Quadrilateral\;TPRQ} \\[3ex] \angle Q + \angle P = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle Q = \angle TQO + \angle OQR ...diagram \\[3ex] \angle Q = \angle TQO + 32 \\[3ex] \angle P = \angle TPQ + \angle QPR ...diagram \\[3ex] \angle P = 15 + 58 \\[3ex] \angle P = 73^\circ \\[3ex] \implies \\[3ex] (\angle TOQ + 32) + 73 = 180 \\[3ex] \angle TOQ + 32 + 73 = 180 \\[3ex] \angle TOQ + 105 = 180 \\[3ex] \angle TOQ = 180 - 105 \\[3ex] \angle TOQ = 75^\circ $(125.) GCSE The lines STR and BCR are tangents to the circle shown. Angle RTC = 47° and angle ADC = 94° John proved that the lines AC and SR are parallel. He used the following proof but didn't give his reasons. Using the properties of tangents and circle theorems complete John's argument. (1.) Angle RCT = 47° because ....... (2.) Angle RTC = Angle TAC because ....... (3.) Angle ATC = 86° because ....... (4.) Angle STA = 47° because ....... (5.) So the lines AC and SR are parallel because ....... (1.) Angle RCT = 47° because: two tangents: STR and BCR drawn from the same external point: R are equal in length. Because |TR| = |CR|, ∠RCT = ∠RTC = 47°...base angles of isosceles triangle TRC (2.) Angle RTC = Angle TAC = 47° because the angle between tangent STR and chord TC = angle in the alternate segment TAC (3.) Angle ATC = 86° because the interior opposite angles of the cyclic quadrilateral ATCD are supplementary$ \angle T + \angle D = 180^\circ ...opposite\;\;\angle s\;\;of\;\;cyclic\;\;quadrilateral\;ATCD\;\;are\;\;supplementary \\[3ex] \angle T = \angle ATC ...diagram \\[3ex] \angle D = \angle ADC = 94^\circ ... diagram \\[3ex] \implies \\[3ex] \angle ATC + 94 = 180 \\[3ex] \angle ATC = 180 - 94 \\[3ex] \angle ATC = 86^\circ \\[3ex] $(4.) Angle STA = 47° because$ \angle STA + \angle ATC + \angle RTC = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle STA + 86 + 47 = 180 \\[3ex] \angle STA + 133 = 180 \\[3ex] \angle STA = 180 - 133 \\[3ex] \angle STA = 47^\circ \\[3ex] $(5.) So the lines AC and SR are parallel because angle STA = angle TAC = 47°...alternate angles are equal Alternate angles between the parallel lines: AC and SR, are equal. (126.) WASSCE In the diagram, O is the centre of the circle,$\angle QPS = 100^\circ$,$\angle PSQ = 60^\circ$and$\angle QSR = 80^\circ$Calculate$\angle SQR A.\;\; 20^\circ \\[3ex] B.\;\; 40^\circ \\[3ex] C.\;\; 60^\circ \\[3ex] D.\;\; 80^\circ \\[3ex]  \underline{\triangle PQS} \\[3ex] \angle PSQ + \angle QPS + \angle SQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQS \\[3ex] 60 + 100 + \angle SQP = 180 \\[3ex] 160 + \angle SQP = 180 \\[3ex] \angle SQP = 180 - 160 \\[3ex] \angle SQP = 20^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \angle Q + \angle S = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle Q = \angle SQP + \angle SQR ...diagram \\[3ex] \angle Q = 20 + \angle SQR \\[3ex] \angle S = 60 + 80 \\[3ex] \angle S = 140^\circ \\[3ex] \implies \\[3ex] 20 + \angle SQR + 140 = 180 \\[3ex] 160 + \angle SQR = 180 \\[3ex] \angle SQR = 180 - 160 \\[3ex] \angle SQR = 20^\circ $(127.) CSEC In the diagram below, not drawn to scale, W, X, Y and Z are points on the circumference of a circle, centre O TYV is a tangent to the circle at Y, ∠XWZ = 64° and ∠ZYV = 23° Calculate, giving reasons for your answer, the measure of angle (i) XYZ (ii) YXZ (iii) OXZ$ (i) \\[3ex] Obtuse\;\;\angle XOZ = 2 * \angle XWZ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle XOZ = 2 * 64 \\[3ex] Obtuse\;\;\angle XOZ = 128^\circ \\[3ex] Reflex\;\;\angle XOZ + Obtuse\;\;\angle XOZ = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle XOZ + 128 = 360 \\[3ex] Reflex\;\;\angle XOZ = 360 - 128 \\[3ex] Reflex\;\;\angle XOZ = 232^\circ \\[3ex] Also: \\[3ex] Reflex\;\;\angle XOZ = 2 * \angle XYZ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 232 = 2 * \angle XYZ \\[3ex] 2 * \angle XYZ = 232 \\[3ex] \angle XYZ = \dfrac{232}{2} \\[5ex] \angle XYZ = 116^\circ \\[3ex] (ii) \\[3ex] 23^\circ = \angle YXZ ...\angle\;\;between\;\;tangent\;TYV\;\;and\;\;chord\;YZ = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle YXZ = 23^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle XOZ} \\[3ex] \angle OXZ = \angle OZX = p ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XOZ \\[3ex] \angle OXZ + \angle OZX + Obtuse\;\;\angle XOZ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XOZ \\[3ex] p + p + 128 = 180 \\[3ex] 2p = 180 - 128 \\[3ex] 2p = 52 \\[3ex] p = \dfrac{52}{2} \\[5ex] p = 26 \\[3ex] \therefore \angle OXZ = \angle OZX = 26^\circ $(128.) WASSCE In the diagram, |ZY| = |XY|, ∠WYZ = 65° and ∠XWY = 48° Find ∠WYX$ A.\;\; 19^\circ \\[3ex] B.\;\; 25^\circ \\[3ex] C.\;\; 45^\circ \\[3ex] D.\;\; 65^\circ \\[3ex]  \angle XZY = \angle XWY = 48^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ZXY = \angle XZY = 48^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XZY \\[3ex] \angle ZXY + \angle XZY + \angle XYZ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XZY \\[3ex] 48 + 48 + \angle XYZ = 180 \\[3ex] 96 + \angle XYZ = 180 \\[3ex] \angle XYZ = 180 - 96 \\[3ex] \angle XYZ = 84^\circ \\[3ex] But: \\[3ex] \angle XYZ = \angle WYX + \angle WYZ ...diagram \\[3ex] 84 = \angle WYX + 65 \\[3ex] \angle WYX + 65 = 84 \\[3ex] \angle WYX = 84 - 65 \\[3ex] \angle WYX = 19^\circ $(129.) WASSCE In the diagram,$\overline{YW}$is a tangent to the circle at X, |UV| = |VX| and ∠VXW = 50° Find the value of ∠UXY$ A.\;\; 70^\circ \\[3ex] B.\;\; 80^\circ \\[3ex] C.\;\; 105^\circ \\[3ex] D.\;\; 110^\circ \\[3ex]  \angle VUX = 50^\circ ...\angle\;\;between\;\;tangent\;YXW\;\;chord\;XV = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle VXU = \angle VUX = 50^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XUV \\[3ex] \angle UXY + \angle VXU + \angle XWY = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle UXY + 50 + 50 = 180 \\[3ex] \angle UXY + 100 = 180 \\[3ex] \angle UXY = 180 - 100 \\[3ex] \angle UXY = 80^\circ $(130.) GCSE In the diagram, O is the centre of the circle. P, Q, R and S are points on the circumference of the circle. ST and QT are tangents to the circle. Angle STQ = x Work out the size of angle SPQ in terms of x Explain each stage of your working clearly.$ \angle OST = 90^\circ \\[3ex] ...radius\;OS \perp tangent\;ST\;\;at\;\;point\;\;of\;\;contact\;S \\[3ex] \angle OQT = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;QT\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] Obtuse\;\;\angle O + \angle OST + \angle T + \angle OQT = 360^\circ \\[3ex] ...sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;quadrilateral\;OSTQ \\[3ex] \implies \\[3ex] Obtuse\;\;\angle O + 90 + x + 90 = 360 \\[3ex] Obtuse\;\;\angle O = 360 - 90 - x - 90 \\[3ex] Obtuse\;\;\angle O = 180 - x \\[3ex] Reflex\;\;\angle O + Obtuse\;\;\angle O = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle O + (180 - x) = 360 \\[3ex] Reflex\;\;\angle O + 180 - x = 360 \\[3ex] Reflex\;\;\angle O = 360 - 180 + x \\[3ex] Reflex\;\;\angle O = 180 + x \\[3ex] Also: \\[3ex] Reflex\;\;\angle O = 2 * \angle SPQ \\[3ex] ...\angle\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 180 + x = 2 * \angle SPQ \\[3ex] 2 * \angle SPQ = 180 + x \\[3ex] \angle SPQ = \dfrac{180 + x}{2} $(131.) GCSE O is the centre of the circle. A, B, C are points on the circumference DB and DC are tangents. The angle CAB is x Find, in simplest form, in terms of x, (a) angle BOC (b) angle BDC Given that the lines AB and CD are parallel, find, in simplest form, in terms of x, (c) angle ABO (d) angle ACO Construction: Join the chord from point B to point C$ (a) \\[3ex] \angle BOC = 2 * \angle CAB ...\angle\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * x \\[3ex] \angle BOC = 2x \\[3ex] (b) \\[3ex] \angle BCD = \angle CAB = x ...\angle\;\;between\;\;tangent\;CD\;\;and\;\;chord\;CB = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] |BD| = |CD|...two\;\;tangents\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point\;D\;\;are\;\;equal\;\;in\;\;length \\[3ex] \angle CBD = \angle BCD = x \;\;base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BDC \\[3ex] \angle CBD + \angle BCD + \angle BDC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BDC \\[3ex] x + x + \angle BDC = 180 \\[3ex] 2x + \angle BDC = 180 \\[3ex] \angle BDC = 180 - 2x \\[3ex]  (c) \\[3ex] \angle OCB = \angle OBC \;\;base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle COB \\[3ex] \angle OCB + \angle OBC + \angle BOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COB \\[3ex] \angle OBC + \angle OBC + 2x = 180 \\[3ex] 2\angle OBC = 180 - 2x \\[3ex] \angle OBC = \dfrac{180 - 2x}{2} \\[5ex] \angle OBC = \dfrac{2(90 - x)}{2} \\[5ex] \angle OBC = 90 - x \\[3ex] \angle ABC = \angle BCD = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Also: \\[3ex] \angle ABC = \angle ABO + \angle OBC ...diagram \\[3ex] \implies \\[3ex] x = \angle ABO + (90 - x) \\[3ex] x = \angle ABO + 90 - x \\[3ex] \angle ABO + 90 - x = x \\[3ex] \angle ABO = x - 90 + x \\[3ex] \angle ABO = 2x - 90 \\[3ex] \angle ABO = 2(x - 45) \\[3ex] (d) \\[3ex] \angle OCE = 90^\circ ...radius\;OC \perp tangent\;EH\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] \angle ACE = \angle CAB = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OCE = \angle ACE + \angle ACO ...diagram \\[3ex] 90 = \angle = x + \angle ACO \\[3ex] x + \angle ACO = 90 \\[3ex] \angle ACO = 90 - x $(132.) WASSCE The diagram shows a circle centre O If$\angle STR = 29^\circ$and$\angle RST = 46^\circ$, calculate the value of$\angle STO A.\;\; 12^\circ \\[3ex] B.\;\; 15^\circ \\[3ex] C.\;\; 29^\circ \\[3ex] D.\;\; 34^\circ \\[3ex]  \angle ROT = 2 * \angle RST...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle ROT = 2 * 46 \\[3ex] \angle ROT = 92^\circ \\[3ex] \angle ORT = \angle OTR = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ROT \\[3ex] \angle ORT + \angle OTR + \angle ROT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROT \\[3ex] p + p + 92 = 180 \\[3ex] 2p = 180 - 92 \\[3ex] 2p = 88 \\[3ex] p = \dfrac{88}{2} \\[5ex] p = 44 \\[3ex] \therefore \angle ORT = \angle OTR = 44^\circ \\[3ex] \angle OTR = \angle STO + \angle STR ...diagram \\[3ex] 44 = \angle STO + 29 \\[3ex] \angle STO + 29 = 44 \\[3ex] \angle STO = 44 - 29 \\[3ex] \angle STO = 15^\circ $(133.) CSEC The diagram below, not drawn to scale, shows a circle, centre O EH and EF are tangents to the circle. FOG and JOH are straight lines. The measure of ∠FEH = 44° Calculate, giving reasons for your answer, the measure of: (i) ∠EHF (ii) ∠FGH (iii) ∠JHE (iv) ∠JGH$ (i) \\[3ex] |EH| = |EF| ...two\;\;tangents\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point\;E\;\;are\;\;equal\;\;in\;\;length \\[3ex] \angle EHF = \angle EFH = p \;\;base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] \angle EHF + \angle EFH + \angle FEH = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FEH \\[3ex] p + p + 44 = 180 \\[3ex] 2p + 44 = 180 \\[3ex] 2p = 180 - 44 \\[3ex] 2p = 136 \\[3ex] p = \dfrac{136}{2} \\[5ex] p = 68 \\[3ex] \therefore \angle EHF = \angle EFH = 68^\circ \\[3ex] (ii) \\[3ex] \angle EHF = \angle FGH = 68^\circ ...\angle\;\;between\;\;tangent\;EH\;\;and\;\;chord\;HF = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] (iii) \\[3ex] \angle JHE = 90^\circ ...radius\;OH \perp tangent\;EH\;\;at\;\;point\;\;of\;\;contact\;H \\[3ex] (iv) \\[3ex] \angle JGH = 90^\circ...\angle\;\;in\;\;a\;\;semicircle $(134.) GCSE The points A, B and C lie on the circumference of a circle, centre O AD is a tangent to the circle. DCB is a straight line. Find the size of each of the following angles, in terms of x Write your answers in their simplest form. (a)$O\hat{C}D$(b)$O\hat{A}B (a) \\[3ex] O\hat{A}D = 90^\circ...radius\;OA \perp tangent\;AD\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \underline{Quadrilateral\;DCOA} \\[3ex] \hat{D} + \hat{C} + \hat{O} + \hat{A} = 360^\circ \\[3ex] ...sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;quadrilateral \\[3ex] \hat{D} = x \\[3ex] \hat{C} = O\hat{C}D \\[3ex] \hat{O} = 2x \\[3ex] \hat{A} = O\hat{A}D = 90^\circ \\[3ex] \implies \\[3ex] x + O\hat{C}D + 2x + 90 = 360 \\[3ex] O\hat{C}D = 360 - 90 - 2x - x \\[3ex] O\hat{C}D = 270 - 3x \\[3ex] (b) \\[3ex] C\hat{O}A = 2 * C\hat{B}A ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2x = 2 * C\hat{B}A \\[3ex] 2 * C\hat{B}A = 2x \\[3ex] C\hat{B}A = \dfrac{2x}{2} \\[5ex] C\hat{B}A = x \\[3ex] \underline{\triangle DBA} \\[3ex] B\hat{D}A + D\hat{B}A + D\hat{A}B = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DBA \\[3ex] B\hat{D}A = x \\[3ex] D\hat{B}A = C\hat{B}A = x \\[3ex] D\hat{A}B = O\hat{A}D + O\hat{A}B ... diagram \\[3ex] D\hat{A}B = 90 + O\hat{A}B \\[3ex] \implies \\[3ex] x + x + (90 + O\hat{A}B) = 180 \\[3ex] 2x + 90 + O\hat{A}B = 180 \\[3ex] O\hat{A}B = 180 - 90 - 2x \\[3ex] O\hat{A}B = 90 - 2x $(135.) WASSCE In the diagram, TS is a tangent to the circle at A AB || CE, ∠AEC = 5x°, ∠ADB = 60° and ∠TAE = x° Find the value of x Let us solve this question in at least two ways First Method: Construction: Draw the chord from point C to point A$ \angle TAE = \angle ECA = x^\circ ...\angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AE = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle ECA = \angle CAB = x^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;\;CE\;\;and\;\;AB\;\;are\;\;equal \\[3ex] \angle BAS = \angle ADB = 60^\circ ... \angle\;\;in\;\;the\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AB \\[3ex] \angle BAS + \angle CAB + \angle CAE + \angle TAE = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 60 + x + \angle CAE + x = 180 \\[3ex] \angle CAE + 60 + 2x = 180 \\[3ex] \angle CAE = 180 - 60 - 2x \\[3ex] \angle CAE = 120 - 2x \\[3ex] \underline{\triangle CAE} \\[3ex] \angle CAE + \angle AEC + \angle ECA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CAE \\[3ex] (120 - 2x) + 5x + x = 180 \\[3ex] 120 - 2x + 6x = 180 \\[3ex] 4x = 180 - 120 \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15^\circ \\[3ex] $OR Second Method: Construction: Draw the straight line EA and extend to K$ \angle KAB = 5x^\circ ... exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;AECB = interior\;\;opposite\;\;\angle \\[3ex] \underline{Straight\;\;Line\;\;KAE} \\[3ex] \angle KAB + \angle BAE = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 5x + \angle BAE = 180 \\[3ex] \angle BAE = 180 - 5x \\[3ex] \angle BAS = \angle ADB = 60^\circ ... \angle\;\;in\;\;the\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AB \\[3ex] \underline{Straight\;\;Line\;\;SAT} \\[3ex] \angle BAS + \angle BAE + \angle TAE = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 60 + (180 - 5x) + x = 180 \\[3ex] 60 + 180 - 5x + x = 180 \\[3ex] 240 - 4x = 180 \\[3ex] 240 - 180 = 4x \\[3ex] 60 = 4x \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15^\circ $(136.) GCSE A, B and C lie on a circle with diameter AC AC is extended to D and angle BAC = 63° Work out angle BCD Give reasons to explain your answer.$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle BCD = \angle BAC + \angle ABC \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle BCD = 63 + 90 \\[3ex] \angle = 153^\circ $(137.) CSEC The diagram below, not drawn to scale, shows a circle, with centre O The points A, B, C and M are on the circumference. The straight line CN is a tangent to the circle at the point C and is perpendicular to BN Determine, giving a reason for your answer,$ (i)\;\; A\hat{B}C \\[3ex] (ii)\;\; C\hat{M}B \\[3ex] (iii)\;\; N\hat{C}M \\[3ex]  (i) \\[3ex] A\hat{C}B = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] C\hat{A}B = 58^\circ ...diagram \\[3ex] C\hat{A}B + A\hat{B}C + A\hat{C}B = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] 58 + A\hat{B}C + 90 = 180 \\[3ex] 148 + A\hat{B}C = 180 \\[3ex] A\hat{B}C = 180 - 148 \\[3ex] A\hat{B}C = 32^\circ \\[3ex] (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABMC} \\[3ex] \angle A + \angle M = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle A = C\hat{A}B = 58^\circ ...diagram \\[3ex] \angle M = C\hat{M}B ...diagram \\[3ex] \implies \\[3ex] 58 + C\hat{M}B = 180 \\[3ex] C\hat{M}B = 180 - 58 \\[3ex] C\hat{M}B = 122^\circ \\[3ex] (iii) \\[3ex] C\hat{M}N = C\hat{A}B = 58^\circ ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;ABMC = interior\;\;opposite\;\;\angle \\[3ex] OR \\[3ex] C\hat{M}N + C\hat{M}B = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] C\hat{M}N + 122 = 180 \\[3ex] C\hat{M}N = 180 - 122 \\[3ex] C\hat{M}N = 58^\circ \\[3ex] \underline{\triangle CMN} \\[3ex] N\hat{C}M + C\hat{M}N + C\hat{N}M = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CMN \\[3ex] N\hat{C}M + 58 + 90 = 180 \\[3ex] N\hat{C}M + 148 = 180 \\[3ex] N\hat{C}M = 180 - 148 \\[3ex] N\hat{C}M = 32^\circ $(138.) curriculum.gov.mt The points B, C, D and E lie on the circumference of a circle. ABC is a straight line where A lies outside the circle. Calculate, giving reasons for your answers: (a) the size of$A\hat{B}E$(b) the size of$B\hat{E}D \underline{Cyclic\;\;Quadrilateral\;DCBE} \\[3ex] (a) \\[3ex] A\hat{B}E = C\hat{D}E = 110^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quad = interior\;\;opposite\;\;\angle \\[3ex] (b) \\[3ex] B\hat{E}D + D\hat{C}B = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] B\hat{E}D + 100 = 180 \\[3ex] B\hat{E}D = 180 - 100 \\[3ex] B\hat{E}D = 80^\circ $(139.) NSC Chord QN bisects$M\hat{N}P$and intersects chord MP at S The tangent at P meets MN produced at R such that QN || PR Let$\hat{P_1} = x$(139.1) Determine the following angles in terms of x. Give reasons.$ (a)\;\; \hat{N_2} \\[3ex] (b)\;\; \hat{Q_2} \\[3ex] $(139.2) Prove, giving reasons, that$\dfrac{MN}{NR} = \dfrac{MS}{SQ} (139.1) \\[3ex] (a) \\[3ex] \hat{N_2} = x...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] (b) \\[3ex] x = M_2...\angle\;\;between\;\;tangent\;PR\;\;and\;\;chord\;PN = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] M_2 = x \\[3ex] Q_2 = M_2 = x ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (139.2) \\[3ex] $To solve this question, knowledge of Triangle Theorems is required.$ Because: QN || PR,\;\; SN\;\;is\;\;also\;\;|| PR...diagram \\[3ex] \dfrac{MN}{NR} = \dfrac{MS}{SP} ...Side\;\;Splitter\;\;theorem \\[5ex] However: \\[3ex] \dfrac{MN}{NR} = \dfrac{MS}{SP} ...Given \\[5ex] So,\;\;let\;\;us\;\;just\;\;show\;\;that\;\;SP = SQ \\[3ex] \underline{\triangle QSP} \\[3ex] \hat{Q_2} = x ...already\;\;shown\;\;in\;\;(b) \\[3ex] \hat{N_1} = \hat{N_2} = x ...chord\;QN\;\;bisects\;\;M\hat{N}P \\[3ex] \hat{P_3} = \hat{N_1} = x ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] Because\;\;\hat{Q_2} = \hat{P_3} = x;\;\;\triangle QSR\;\;is\;\;isosceles \\[3ex] \implies \\[3ex] SQ = SP \\[3ex] \therefore \dfrac{MN}{NR} = \dfrac{MS}{SP} $(140.) GCSE The points A, B, C and D lie on a circle. AC is a diameter of the circle. ST is the tangent to the circle at A Find the value of (i) x (ii) y$ (i) \\[3ex] AC = diameter \\[3ex] \dfrac{AC}{2} = radius \\[5ex] \angle OAS = 90^\circ \\[3ex] ...radius \perp tangent\;SAT\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAS = \angle SAB + \angle BAO ... diagram \\[3ex] 90 = 48 + \angle BAO \\[3ex] \angle BAO + 48 = 90 \\[3ex] \angle BAO = 90 - 48 \\[3ex] \angle BAO = 42^\circ \\[3ex] x = \angle BAO = 42^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \angle ACD = \angle ABD = 27^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle COD} \\[3ex] \angle OCD + \angle ODC + \angle COD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COD \\[3ex] \angle OCD = \angle ACD = 27^\circ ... diagram \\[3ex] \angle ODC = x^\circ \\[3ex] \angle COD = y^\circ \\[3ex] \implies \\[3ex] 27 + x + y = 180 \\[3ex] 27 + 42 + y = 180 \\[3ex] 69 + y = 180 \\[3ex] y = 180 - 69 \\[3ex] y = 111^\circ $(141.) CSEC In the diagram below, E, C, G and F are points on the circumference of a circle. EG is a diameter of the circle. The tangent AEB is parallel to CD Angle AEC = 68° and angle EFD = 106° Determine the value of EACH of the following angles. Show detailed working where necessary and give a reason to support your answer. (i) ECD (ii) CEG (iii) CGF$ (i) \\[3ex] \angle ECD = 68^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \angle OEA = 90^\circ ...radius\;OE \perp tangent\;AEB\;\;at\;\;point\;\;of\;\;contact\;E \\[3ex] \angle OEA = \angle AEC + \angle CEG \\[3ex] 90 = 68 + \angle CEG \\[3ex] 68 + \angle CEG = 90 \\[3ex] \angle CEG = 90 - 68 \\[3ex] \angle CEG = 22^\circ \\[3ex] (iii) \\[3ex] \angle EFD = \angle ECD + \angle CEF ...exterior\;\;\angle\;\;of\;\;\triangle CEF = sum\;\;of\;\;the\;\;two\;\;interior\;\;\angle s \\[3ex] 106 = 68 + \angle CEF \\[3ex] 68 + \angle CEF = 106 \\[3ex] \angle CEF = 106 - 68 \\[3ex] \angle CEF = 38^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ECGF} \\[3ex] \angle G + \angle E = 180^\circ ...interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral\;\;are\;\;supplementary \\[3ex] \angle G = \angle CGF ...diagram \\[3ex] \angle E = \angle CEF ... diagram \\[3ex] \implies \\[3ex] \angle CGF + \angle CEF = 180 \\[3ex] \angle CGF + 38 = 180 \\[3ex] \angle CGF = 180 - 38 \\[3ex] \angle CGF = 142^\circ $(142.) GCSE A, B, C and D lie on a circle centre O Angle ABC = 58° and angle CAD = 23° Calculate (a) angle OCA (b) angle DCA$ (a) \\[3ex] Obtuse\;\;\angle COA = 2 * \angle CBA \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle COA = 2 * 58 \\[3ex] Obtuse\;\;\angle COA = 116^\circ \\[3ex] \angle OCA = \angle OAC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle COA \\[3ex] \angle OCA + \angle OAC + \angle COA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COA \\[3ex] p + p + 116 = 180 \\[3ex] 2p = 180 - 116 \\[3ex] 2p = 64 \\[3ex] p = \dfrac{64}{2} \\[5ex] p = 32 \\[3ex] \therefore \angle OCA = \angle OAC = 32^\circ \\[3ex] (b) \\[3ex] Reflex\;\;\angle COA + Obtuse\;\;\angle COA = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle COA + 116 = 360 \\[3ex] Reflex\;\;\angle COA = 360 - 116 \\[3ex] Reflex\;\;\angle COA = 244^\circ \\[3ex] Also: \\[3ex] Reflex\;\;\angle COA = 2 * \angle CDA \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 244 = 2 * \angle CDA \\[3ex] 2 * \angle CDA = 244 \\[3ex] \angle CDA = \dfrac{244}{2} \\[5ex] \angle CDA = 122^\circ \\[3ex] \angle DCA + \angle CAD + \angle CDA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CDA \\[3ex] \angle DCA + 23 + 122 = 180 \\[3ex] \angle DCA + 145 = 180 \\[3ex] \angle DCA = 180 - 145 \\[3ex] \angle DCA = 35^\circ $WASSCE In the diagram, O is the centre of the circle, RT is a diameter, ∠PQT = 33° and ∠TOS = 76° Use this diagram to answer questions 143 and 144 (143.) WASSCE Calculate the value of ∠PTR$ A.\;\; 73^\circ \\[3ex] B.\;\; 67^\circ \\[3ex] C.\;\; 57^\circ \\[3ex] D.\;\; 37^\circ \\[3ex]  \angle PRT = 33^\circ...\angle s\;\;in\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle TPR = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PTR + \angle TPR + \angle PRT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TPR \\[3ex] \angle PTR + 90 + 33 = 180 \\[3ex] \angle PTR + 123 = 180 \\[3ex] \angle PTR = 180 - 123 \\[3ex] \angle PTR = 57^\circ $(144.) WASSCE Find the size of ∠PRS$ A.\;\; 76^\circ \\[3ex] B.\;\; 71^\circ \\[3ex] C.\;\; 38^\circ \\[3ex] D.\;\; 33^\circ \\[3ex]  \angle OTS = \angle OST = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle TOS \\[3ex] \angle OTS + \angle OST + \angle TOS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TOS \\[3ex] k + k + 76 = 180 \\[3ex] 2k = 180 - 76 \\[3ex] 2k = 104 \\[3ex] k = \dfrac{104}{2} \\[5ex] k = 52 \\[3ex] \therefore \angle OTS = \angle OST = 52^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PRST} \\[3ex] \angle T + \angle R = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle T = \angle PTR + \angle OTS ...diagram \\[3ex] \angle T = 57 + 52 \\[3ex] \angle T = 109^\circ \\[3ex] \angle PRS = \angle R ...diagram \\[3ex] \implies \\[3ex] 109 + \angle PRS = 180 \\[3ex] \angle PRS = 180 - 109 \\[3ex] \angle PRS = 71^\circ $(145.) GCSE The diagram shows a circle with centre O The three points A, B and C lie on the circumference of the circle. The tangent XAY touches the circle at A Given that$A\hat{O}C = 6x^\circ$, find an expression for$Y\hat{A}C$in terms of x You must give a reason for each stage of your working. Express your answer in simplest form.$ A\hat{O}C = 2 * A\hat{B}C ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 6x = 2 * A\hat{B}C \\[3ex] 2 * A\hat{B}C = 6x \\[3ex] A\hat{B}C = \dfrac{6x}{2} \\[5ex] A\hat{B}C = 3x \\[3ex] Y\hat{A}C = A\hat{B}C = 3x ...\angle \:\:between\:\:tangent\;XAY\;\;and\:\:chord\;AC = \angle\:\:in\:\:the\;\;alternate\:\:segment $(146.) WASSCE The diagram shows a circle ABCD with centre O and radius 7cm The reflex angle AOC = 190 degrees and angle DAO = 35 degrees Find: (i) ∠ABC (ii) ∠ADC$ (i) \\[3ex] Reflex\;\;\angle AOC = 2 * \angle ABC \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 190 = 2 * \angle ABC \\[3ex] 2 * \angle ABC = 190 \\[3ex] \angle ABC = \dfrac{190}{2} \\[5ex] \angle ABC = 95^\circ \\[3ex] (ii) \\[3ex] Reflex\;\;\angle AOC + Obtuse\;\;\angle AOC = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] 190 + Obtuse\;\;\angle AOC = 360 \\[3ex] Obtuse\;\;\angle AOC = 360 - 190 \\[3ex] Obtuse\;\;\angle AOC = 170^\circ \\[3ex] Also: \\[3ex] Obtuse\;\;\angle AOC = 2 * \angle ADC \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 170 = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 170 \\[3ex] \angle ADC = \dfrac{170}{2} \\[5ex] \angle ADC = 85^\circ $(147.) WASSCE In the diagram,$\overline{TU}$is a tangent to the circle. ∠RVU = 100° and ∠URS = 36° Calculate the value of angle STU$ \angle SUT = \angle URS = 36^\circ ...\angle \:\:between\:\:tangent\;TU\;\;and\:\:chord\;US = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] \angle UST = \angle V = 100^\circ ...exterior\;\;\angle\;\;of\;\;Cyclic\;\;Quadrilateral\;UVRS = interior\;\;opposite\;\;\angle \\[3ex] \underline{\triangle UST} \\[3ex] \angle STU + \angle UST + \angle SUT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TOS \\[3ex] \angle STU + 100 + 36 = 180 \\[3ex] \angle STU + 136 = 180 \\[3ex] \angle STU = 180 - 136 \\[3ex] \angle STU = 44^\circ $(148.) GCSE A, B, P and Q lie on the circle, centre O Angle APB = 56° Find the value of (a) x (b) y$ (a) \\[3ex] x = 2 * 56 ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] x = 112^\circ \\[3ex] (b) \\[3ex] y = 56^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $(149.) curriculum.gov.mt AC and BC are tangents to a circle centre O. a) What is the value of the angle CAO? Give a reason for your answer. b) Explain why triangles AOC and BOC are congruent. c) What can you say about the tangents AC and BC?$ a) \\[3ex] \angle CAO = 90^\circ ... radius\;OA \perp tangent\;AC \;\;at\;\;point\;\;of\;\;contact\; A \\[5ex] b) \\[3ex] At\;\;least\;\;three\;\;ways\;\;to\;\;show\;\;Congruency \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Angle:\;\; \angle OAC = \angle OBC = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AC = BC ... ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \triangle AOC \cong \triangle BOC...Side-Angle-Side \\[3ex] OR \\[3ex] Angle:\;\; \angle OAC = \angle OBC = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AC = BC ... ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] Angle:\;\; \angle ACO = \angle BCO ... \\[3ex] $For two tangents: AC and BC drawn from the same external point: C, the line joining the external point: C, and the centre of the circle: O; bisects the angle between the tangents.$ \implies \\[3ex] \triangle AOC \cong \triangle BOC...Angle-Side-Angle \\[3ex] OR \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Side:\;\; AC = BC ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] Side:\;\; OC = OC ... common\;\;side\;\;of\;\;the\;\;two\;\;\triangle s \\[3ex] \implies \\[3ex] \triangle AOC \cong \triangle BOC...Side-Side-Side \\[5ex] c) \\[3ex] AC = BC ... ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length $(150.) NZQA For the diagram above: (i) Find the size of angle reflex COA, x Explain your reasoning (ii) Find the size of angle DCO, y Give geometric reasons for each step in your solution.$ (i) \\[3ex] Obtuse\;\;\angle COA = 2 * \angle CDA \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle COA = 2 * 72 \\[3ex] Obtuse\;\;\angle COA = 144^\circ \\[3ex] Reflex\;\;\angle COA + Obtuse\;\;\angle COA = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] x + 144 = 360 \\[3ex] x = 360 - 144 \\[3ex] x = 216^\circ \\[3ex] $Construction: Join the chord from point C to point A$ (ii) \\[3ex] \angle OCA = \angle OAC = p \\[3ex] ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOC \\[3ex] \angle OCA + \angle OAC + Reflex\;\;\angle COA = 180^\circ \\[3ex] ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle COA \\[3ex] \implies \\[3ex] p + p + 144 = 180 \\[3ex] 2p = 180 - 144 \\[3ex] 2p = 36 \\[3ex] p = \dfrac{36}{2} \\[5ex] p = 18^\circ \\[3ex] \therefore \angle OCA = \angle OAC = 18^\circ \\[3ex] \underline{\triangle DCA} \\[3ex] \angle DCA + \angle CAD + \angle ADC = 180^\circ \\[3ex] ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle COA \\[3ex] \angle DCA = \angle DCO + \angle OCA ...diagram \\[3ex] \angle DCA = y + 18 \\[3ex] \angle CAD = \angle OAC + \angle OAA ...diagram \\[3ex] \angle CAD = 18 + 38 \\[3ex] \angle CAD = 56^\circ \\[3ex] \angle ADC = 72^\circ ...diagram \\[3ex] \implies \\[3ex] (y + 18) + 56 + 72 = 180 \\[3ex] y + 18 + 56 + 72 = 180 \\[3ex] y + 146 = 180 \\[3ex] y = 180 - 146 \\[3ex] y = 34^\circ $(151.) GCSE O is the centre of the circle and the tangent TW touches the circle at B Find the size of the angles (a) TBO (b) CAB (c) CBW (d) DBC$ (a) \\[3ex] \angle TBO = 90^\circ ... radius\;OB \perp tangent\;TBW\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] (b) \\[3ex] \underline{\triangle BOC} \\[3ex] \angle OBC = \angle OCB = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOC \\[3ex] \angle OBC + \angle OCB + \angle BOC = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle BOC \\[3ex] \implies \\[3ex] p + p + 72 = 180 \\[3ex] 2p = 180 - 72 \\[3ex] 2p = 108 \\[3ex] p = \dfrac{108}{2} \\[5ex] p = 54 \\[3ex] \therefore \angle OBC = \angle OCB = 54^\circ \\[3ex] \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle ABC \\[3ex] \angle ACB = \angle OCB = 54^\circ \\[3ex] \implies \\[3ex] \angle CAB + 90 + 54 = 180 \\[3ex] \angle CAB + 144 = 180 \\[3ex] \angle CAB = 180 - 144 \\[3ex] \angle CAB = 36^\circ \\[3ex] (c) \\[3ex] \angle CBW = \angle CAB = 36^\circ ...\angle \:\:between\:\:tangent\;TBW\;\;and\:\:chord\;BC = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] (d) \\[3ex] \angle DBC = \angle OBC = 54^\circ ...diagram $(152.) WASSCE In the diagram, PQ and PS are tangents to the circle centre O If ∠PSQ = m, ∠SPQ = n and ∠SQR = 33°, find the value of (m + n)$ A.\;\; 103^\circ \\[3ex] B.\;\; 123^\circ \\[3ex] C.\;\; 133^\circ \\[3ex] D.\;\; 143^\circ \\[3ex]  \angle RQK = 90^\circ \\[3ex] ... radius\;OQ \perp tangent\;PQK\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle SQK = \angle SQR + \angle RQK ...diagram \\[3ex] \angle SQK = 33 + 90 \\[3ex] \angle SQK = 123^\circ \\[3ex] \angle SQK = m + n \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] m + n = \angle SQK \\[3ex] \therefore m + n = 123^\circ \\[3ex] $OR$ \angle RQP = 90^\circ \\[3ex] ... radius\;OQ \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] Also: \\[3ex] \angle RQP = \angle SQR + \angle SQP ...diagram \\[3ex] 90 = 33 + \angle SQP \\[3ex] 33 + \angle SQP = 90 \\[3ex] \angle SQP = 90 - 33 \\[3ex] \angle SQP = 57^\circ \\[3ex] \underline{\triangle PSQ} \\[3ex] \angle SPQ + \angle PSQ + \angle SQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] m + n + 57 = 180 \\[3ex] m + n = 180 - 57 \\[3ex] m + n = 123^\circ $(153.) curriculum.gov.mt a) AP is a tangent to circle O and ABCD is a cyclic quadrilateral. Work out the values of angles a, b, and c, giving reasons for your answers. b) AC is a tangent to circle centre O. In$\triangle AOC$, angles x and y are in ratio 1:2 Work out the value of angle z$ a) \\[3ex] \angle DAB = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] a + \angle DAB + 65 = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] a + 90 + 65 = 180 \\[3ex] a + 155 = 180 \\[3ex] a = 180 - 155 \\[3ex] a = 25^\circ \\[3ex] 65^\circ = b ...\angle\;\;between\;\;tangent\;AP\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] b = 65^\circ \\[3ex] c + (b + 72) = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] c + (65 + 72) = 180 \\[3ex] c + 137 = 180 \\[3ex] c = 180 - 137 \\[3ex] c = 43^\circ \\[3ex] OR \\[3ex] a = \angle DBA ...\angle\;\;between\;\;tangent\;AP\;\;and\;\;chord\;AD = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DBA = a = 25^\circ \\[3ex] \angle DCB = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CBD + \angle DCB + \angle CDB = 180^\circ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle DCB \\[3ex] \angle CBD + 90 + 72 = 180 \\[3ex] \angle CBD + 162 = 180 \\[3ex] \angle CBD = 180 - 162 \\[3ex] \angle CBD = 18^\circ \\[3ex] c = \angle CBD + \angle DBA ... diagram \\[3ex] c = 18 + 25 \\[3ex] c = 43^\circ \\[5ex] b) \\[3ex] x:y = 1:2 \\[3ex] \implies \\[3ex] \dfrac{x}{y} = \dfrac{1}{2} \\[5ex] y(1) = x(2) \\[3ex] y = 2x ... eqn.(1) \\[3ex] \angle OAC = 90^\circ ... radius\;OA \perp tangent\;AC \;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAC + x + y = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAC \\[3ex] 90 + x + y = 180 \\[3ex] x + y = 180 - 90 \\[3ex] x + y = 90 \\[3ex] But:\;\; y = 2x...eqn.(1) \\[3ex] x + 2x = 90 \\[3ex] 3x = 90 \\[3ex] x = \dfrac{90}{3} \\[5ex] x = 30^\circ \\[3ex] From\;\;eqn.(1):\;\;y = 2x \\[3ex] y = 2(30) \\[3ex] y = 60^\circ \\[3ex] y = 2z ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2z = y \\[3ex] z = \dfrac{y}{2} \\[5ex] z = \dfrac{60}{2} \\[5ex] z = 30^\circ $(154.) WASSCE The diagram is a circle passing through the points A, B, C and D such that AC and BD meet at a point E inside the circle. If ∠DAC = 27°, ∠ABD = 54° and ∠ACB = 63°, find: (i) ∠CAB (ii) ∠AEB$ (i) \\[3ex] \angle DBC = \angle DAC = 27^\circ \\[3ex] ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ABC = \angle ABD + \angle DBC ...diagram \\[3ex] \angle ABC = 54 + 27 \\[3ex] \angle ABC = 81^\circ \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle CAB + 81 + 63 = 180 \\[3ex] \angle CAB + 144 = 180 \\[3ex] \angle CAB = 180 - 144 \\[3ex] \angle CAB = 36^\circ \\[3ex] (ii) \\[3ex] \angle ADB = \angle ACB = 63^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle AEB = \angle DAC + \angle ADB \\[3ex] ... exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle AEB = 27 + 63 \\[3ex] \angle AEB = 90^\circ \\[3ex] OR \\[3ex] \underline{\triangle AEB} \\[3ex] \angle AEB + \angle EAB + \angle ABE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AEB \\[3ex] \angle EAB = \angle CAB = 36^\circ ...diagram \\[3ex] \angle ABE = \angle ABD = 54^\circ ...diagram \\[3ex] \implies \\[3ex] \angle AEB + 36 + 54 = 180 \\[3ex] \angle AEB + 90 = 180 \\[3ex] \angle AEB = 180 - 90 \\[3ex] \angle AEB = 90^\circ $(155.) NZQA Nails P, Q, R, and W all lie on the circumference of a circle, centre C Lines QRS and PWS are both straight. Find the size, z, of angle RSW, in terms of x and y Justify your answer with clear geometric reasoning$ \angle PWR + y = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle PWR = 180 - y \\[3ex] \underline{\triangle PRW} \\[3ex] \angle PRW = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle RPW + \angle PRW + \angle PWR = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRW \\[3ex] \angle RPW + 90 + (180 - y) = 180 \\[3ex] \angle RPW + 90 + 180 - y = 180 \\[3ex] \angle RPW = 180 - 90 - 180 + y \\[3ex] \angle RPW = y - 90 \\[3ex] \angle RPS = \angle RPW = y - 90 ... diagram \\[3ex] \angle QPR = x ... diagram \\[3ex] \angle QPS = \angle QPR + \angle RPS...diagram \\[3ex] \angle QPS = x + (y - 90) \\[3ex] \angle QPS = x + y - 90 \\[3ex] \angle QSP = z ...diagram \\[3ex] y = \angle Q ...exterior\;\;\angle\;\;of\;\;Cyclic\;\;Quad\;\;PQRW = interior\;\;opposite\;\;\angle \\[3ex] \underline{\triangle PQS} \\[3ex] \angle PQS = \angle Q = y ...diagram \\[3ex] \angle QPS + \angle PQS + \angle QSP = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQS \\[3ex] (x + y - 90) + y + z = 180 \\[3ex] x + y - 90 + y + z = 180 \\[3ex] x + 2y - 90 + z = 180 \\[3ex] z = 180 - x - 2y + 90 \\[3ex] z = 270 - x - 2y $(156.) GCSE AC is a diameter of the circle, centre O Calculate angle ACB$ \angle ABC = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle BAC + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] 34 + 90 + \angle ACB = 180 \\[3ex] 124 + \angle ACB = 180 \\[3ex] \angle ACB = 180 - 124 \\[3ex] \angle ACB = 56^\circ $(157.) SQA National 5 Maths The diagram below shows a circle, centre O. AB and CB are tangents to the circle. AC and ED are parallel. Angle AOD is 143° Calculate the size of angle ABC.$ \angle EOA + \angle AOD = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle EOA + 143 = 180 \\[3ex] \angle EOA = 180 - 143 \\[3ex] \angle EOA = 37^\circ \\[3ex] \angle CAO = \angle EOA = 37^\circ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OAB = 90^\circ...radius\;OA\;\;\perp\;\;tangent\;AB\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAB = \angle CAO + \angle CAB ...diagram \\[3ex] 90 = 37 + \angle CAB \\[3ex] \angle CAB + 37 = 90 \\[3ex] \angle CAB = 90 - 37 \\[3ex] \angle CAB = 53^\circ \\[3ex]  \angle OCA = \angle CAO = 37^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OCB = 90^\circ...radius\;OA\;\;\perp\;\;tangent\;AB\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OCB = \angle OCA + \angle ACB ...diagram \\[3ex] 90 = 37 + \angle ACB \\[3ex] \angle ACB + 37 = 90 \\[3ex] \angle ACB = 90 - 37 \\[3ex] \angle ACB = 53^\circ \\[3ex] \underline{\triangle BAC} \\[3ex] \angle ABC + \angle CAB + \angle ACB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle ABC + 53 + 53 = 180 \\[3ex] \angle ABC + 106 = 180 \\[3ex] \angle ABC = 180 - 106 \\[3ex] \angle ABC = 74^\circ $(158.) WASSCE In the diagram, the two circles intersect at X and Y The centre of the smaller circle is on the circumference of the bigger circle. A and B are any two points on the major arcs, one on each circle. Find an equation connecting a and b Construction: Draw two radii: one radius from point X to point O and the other radius from point Y to point O$ \angle XOY = 2 * \angle XBY ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle XOY = 2 * b \\[3ex] \angle XOY = 2b \\[3ex] \underline{Cyclic\;\;Quadrilateral\;AXOY} \\[3ex] \angle A + \angle O = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A = a \\[3ex] \angle O = \angle XOY = 2b \\[3ex] \implies \\[3ex] a + 2b = 180 $(159.) GCSE The points A, B, C and D lie on the circumference of a circle, centre O EF is a tangent to the circle at C AB = AC$B\hat{C}E = 38^\circ$and$A\hat{C}D = 41^\circ$Write down the size of$ (a)\;\; B\hat{A}C \\[3ex] (b)\;\; A\hat{B}C \\[3ex] (c)\;\; A\hat{D}C \\[3ex] (d)\;\; C\hat{O}B \\[3ex]  (a) \\[3ex] B\hat{A}C = 38^\circ ...\angle \:\:between\:\:tangent\;ECF\;\;and\:\:chord\;BC = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] (b) \\[3ex] A\hat{B}C = A\hat{C}B = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BAC \\[3ex] A\hat{B}C + A\hat{C}B + B\hat{A}C = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle BAC \\[3ex] \implies \\[3ex] p + p + 38 = 180 \\[3ex] 2p = 180 - 38 \\[3ex] 2p = 142 \\[3ex] p = \dfrac{142}{2} \\[5ex] p = 71 \\[3ex] \therefore A\hat{B}C = A\hat{C}B = 71^\circ \\[3ex] (c) \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:ABCD} \\[3ex] \hat{D} + \hat{B} = 180^\circ ...interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral\;\;are\;\;supplementary \\[3ex] \hat{D} = A\hat{D}C ...diagram \\[3ex] \hat{B} = A\hat{B}C = 71^\circ ...diagram \\[3ex] \implies \\[3ex] A\hat{D}C + 71 = 180 \\[3ex] A\hat{D}C = 180 - 71 \\[3ex] A\hat{D}C = 109^\circ \\[3ex]  (d) \\[3ex] C\hat{O}B = 2 * B\hat{A}C ...\angle\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] C\hat{O}B = 2 * 38 \\[3ex] C\hat{O}B = 76^\circ $(160.) NZQA EFGD is a cyclic quadrilateral with angle EDG = 82° O is the centre of the circle. Find the size of angle HFJ Explain your geometric reasoning clearly and logically.$ \underline{Cyclic\;\;Quadrilateral\;\;DEFG} \\[3ex] \angle EDG + \angle EFG = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 82 + \angle EFG = 180 \\[3ex] \angle EFG = 180 - 82 \\[3ex] \angle EFG = 98^\circ \\[3ex] \angle HFJ = \angle EFG = 98^\circ ...vertical\;\;\angle s\;\;are\;\;equal $(161.) curriculum.gov.mt The above diagram shows a circle centre O where A, B, C and D are four points on the circle. TD is a tangent touching the circle at D. Calculate the angles a, b, c, d and e. Give reasons and show your working where necessary.$ a = 40^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[5ex] a + b = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] 40 + b = 90 \\[3ex] b = 90 - 40 \\[3ex] b = 50^\circ \\[5ex] (c + 40) + 115 = 180^\circ ...opposite\;\;\angle s\;\;of\;\;cyclic\;\;quadrilateral\;ABCD\;\;are\;\;supplementary \\[3ex] c + 40 + 115 = 180 \\[3ex] c + 155 = 180 \\[3ex] c = 180 - 155 \\[3ex] c = 25^\circ \\[5ex] d = 40^\circ ...\angle\;\;between\;\;tangent\;TD\;\;and\;\;chord\;CD = \angle\;\;in\;\;alternate\;\;segment \\[5ex] \angle ADC = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle ADT} \\[3ex] 40 + (\angle ADC + d) + e = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle ADT \\[3ex] 40 + (90 + 40) + e = 180 \\[3ex] 40 + 90 + 40 + e = 180 \\[3ex] 170 + e = 180 \\[3ex] e = 180 - 170 \\[3ex] e = 10^\circ $(162.) GCSE The diagram shows a circle, centre O P, Q and R are points on the circumference PQ = 17 cm and QR = 9 cm (a) Explain why angle PQR is 90° (b) Calculate the length PR$ (a) \\[3ex] \angle PQR = 90^\circ...\angle\;\;in\;\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] (b) \\[3ex] |PR|^2 = |PQ|^2 + |QR|^2 ...Pythagorean\;\;theorem \\[3ex] |PR|^2 = 17^2 + 9^2 \\[3ex] |PR|^2 = 289 + 81 \\[3ex] |PR|^2 = 370 \\[3ex] |PR| = \sqrt{370} \\[3ex] |PR| = 19.23538406\;cm $(163.) JAMB TQ is tangent to circle XYTR$\angle YXT = 32^\circ$,$\angle RTQ = 40^\circ$Find$\angle YTR A.\;\; 108^\circ \\[3ex] B.\;\; 121^\circ \\[3ex] C.\;\; 140^\circ \\[3ex] D.\;\; 148^\circ \\[3ex]  \angle\;\;in\;\;alternate\;\;segment = 32^\circ \\[3ex] \angle \;\;between\;\;chord\;YT\;\;and\;\;tangent\;TQ = \theta \\[3ex] \theta = 32^\circ ...\angle \;\;between\;\;chord\;YT\;\;and\;\;tangent\;TQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle YTR = \phi \\[3ex] \theta + \phi + 40 = 180^\circ ...\angle s\;\;in\;\;a\;\;straight\;\;line \\[3ex] 32 + \phi + 40 = 180 \\[3ex] \phi = 180 - 72 \\[3ex] \phi = 108 \\[3ex] \therefore \angle YTR = 108^\circ \\[3ex] OR \\[3ex] \angle TXR = 40^\circ ...\angle \;\;between\;\;tangent\;TQ\;\;and\;\;chord\;TR = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle YXR = 32 + 40 ...diagram \\[3ex] \angle YXR = 72^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;YXRT} \\[3ex] \angle YXR + \angle YTR = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 72 + \angle YTR = 180 \\[3ex] \angle YTR = 180 - 72 \\[3ex] \angle YTR = 108^\circ $(164.) curriculum.gov.mt A, B, and C lie on the circumference of a circle centre O D is the mid-point of AB and CD passes through O (a) What is the size of$A\hat{D}C$? Give a reason for your answer. (b) Show that triangles ADC and BDC are congruent.$ (a) \\[3ex] A\hat{D}C = 90^\circ \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] (b) \\[3ex] \underline{\triangle ADC} \\[3ex] |CD| = diameter...Side \\[3ex] A\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] |AD| = |DB| ...D\;\;is\;\;the\;\;mid-point...Side \\[3ex] \underline{\triangle BDC} \\[3ex] |CD| = diameter...Side \\[3ex] B\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] |DB| = |AD| ...D\;\;is\;\;the\;\;mid-point...Side \\[3ex] \therefore \triangle ADC \cong \triangle BDC ...Side-Angle-Side \\[5ex] OR \\[3ex] \underline{\triangle ADC} \\[3ex] A\hat{C}D = B\hat{C}D ...Angle \\[3ex] ...diameter\;CD\;that\;\;bisects\;\;chord\;AB\;\;also\;\;bisects\;\;its\;\;arc \\[3ex] |CD| = diameter...Side \\[3ex] A\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] \underline{\triangle BDC} \\[3ex] B\hat{C}D = A\hat{C}D ...Angle \\[3ex] ...diameter\;CD\;that\;\;bisects\;\;chord\;AB\;\;also\;\;bisects\;\;its\;\;arc \\[3ex] |CD| = diameter...Side \\[3ex] B\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] \therefore \triangle ADC \cong \triangle BDC ...Angle-Side-Angle $(165.) curriculum.gov.mt In the diagram below, O is the centre of the circle and PQ is a tangent to the circle at A. Find the value of angles a, b, c and d. Give reasons for all your workings.$ a = 60^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[5ex] a + b = 90^\circ ... radius\;OA \perp tangent\;PAQ \;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] b = 90 - a \\[3ex] b = 90 - 60 \\[3ex] b = 30^\circ \\[5ex] b = c ...\angle\;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] c = b \\[3ex] c = 30^\circ \\[5ex] 110^\circ = d + 60^\circ ...exterior\;\;\angle\;\;of\;\;\triangle BCD = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] d + 60 = 110 \\[3ex] d = 110 - 60 \\[3ex] d = 50^\circ $(166.) NZQA The points A, Q, Z, N lie on the circumference of a circle centre O AQ is parallel to NZ Find the size of angle NZQ, in terms of x Explain your geometric reasoning clearly and logically Construction: Extend the two parallel lines: AQ and NZ$ \angle BAN = \angle ANZ = x^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle NZQ = \angle BAN = x^\circ \\[3ex] ...exterior\:\: \angle \:\:of\:\:a\:\:cyclic\:\:quad = interior\:\:opposite\:\: \angle $(167.) curriculum.gov.mt AP and BP are two tangents to a circle centre O. (a) Write down the size of angle OAP. (b) Prove that triangles AOP and BOP are congruent. Construction: Join the line from the radius of the circle: O to the point: P.$ (a) \\[3ex] \angle OAP = 90^\circ ... radius\;OA \perp tangent\;AC \;\;at\;\;point\;\;of\;\;contact\; A \\[5ex] (b) \\[3ex] At\;\;least\;\;three\;\;ways\;\;to\;\;prove\;\;Congruency \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Angle:\;\; \angle OBP = \angle OAP = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AP = BP ... ...two\;\;tangents:\;\;AP\;\;and\;\;BP\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \triangle AOP \cong \triangle BOP...Side-Angle-Side \\[3ex] OR \\[3ex] Angle:\;\; \angle OBP = \angle OAP = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AP = BP ... ...two\;\;tangents:\;\;AP\;\;and\;\;BP\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] Angle:\;\; \angle APO = \angle BPO ... \\[3ex] $For two tangents: AP and BP drawn from the same external point: P, the line joining the external point: P, and the centre of the circle: O; bisects the angle between the tangents.$ \implies \\[3ex] \triangle AOP \cong \triangle BOP...Angle-Side-Angle \\[3ex] OR \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Side:\;\; AP = BP ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] Side:\;\; OP = OP ... common\;\;side\;\;of\;\;the\;\;two\;\;\triangle s \\[3ex] \implies \\[3ex] \triangle AOP \cong \triangle BOP...Side-Side-Side $(168.) GCSE The points A, B, C and D lie on the circumference of a circle, centre O, and AB = AD Prove that y = 2x You must include written reasons in your answer.$ \angle BDA = \angle DBA = x...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DBA \\[3ex] \angle BDA + \angle DBA + \angle DAB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DBA \\[3ex] \implies \\[3ex] x + x + \angle DAB = 180 \\[3ex] 2x + \angle DAB = 180 \\[3ex] \angle DAB = 180 - 2x \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:CBAD} \\[3ex] \angle C + \angle A = 180^\circ \\[3ex] ...interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A = \angle DAB = 180 - 2x...diagram \\[3ex] \angle C = y^\circ...diagram \\[3ex] \implies \\[3ex] y + (180 - 2x) = 180 \\[3ex] y + 180 - 2x = 180 \\[3ex] y = 180 - 180 + 2x \\[3ex] y = 2x $(169.) JAMB From the cyclic quadrilateral MNOP above, find the value of$x A.\;\; 16^\circ \\[3ex] B.\;\; 25^\circ \\[3ex] C.\;\; 42^\circ \\[3ex] D.\;\; 39^\circ \\[3ex]  (2x + 18) + 84 = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 2x + 18 + 84 = 180 \\[3ex] 2x + 102 = 180 \\[3ex] 2x = 180 - 102 \\[3ex] 2x = 78 \\[3ex] x = \dfrac{78}{2} \\[5ex] x = 39^\circ $(170.) GCSE The points A, B, C and D lie on the circumference of a circle with centre O All the angles are measured in degrees. Find the size of the obtuse angle$A\hat{O}C Obtuse\;\;A\hat{O}C = 2 * A\hat{B}C \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;A\hat{O}C = 2 * 2x \\[3ex] Obtuse\;\;A\hat{O}C = 4x \\[3ex] Reflex\;\;A\hat{O}C = 2 * A\hat{D}C \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;A\hat{O}C = 2 * 3x \\[3ex] Reflex\;\;A\hat{O}C = 6x \\[3ex] Reflex\;\;A\hat{O}C + Obtuse\;\;A\hat{O}C = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] 6x + 4x = 130 \\[3ex] 10x = 130 \\[3ex] x = \dfrac{130}{10} \\[5ex] x = 13 \\[3ex] Obtuse\;\;A\hat{O}C \\[3ex] = 4x \\[3ex] = 4(13) \\[3ex] = 52^\circ $(171.) NZQA (i) Show that angle j = 120° Explain your method clearly, and give geometric reasons for each step. (ii) Find the sizes of angle h and angle k$ (i) \\[3ex] x + 120 = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] x = 360 - 120 \\[3ex] x = 240^\circ \\[3ex] x = 2 * j ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2j = x \\[3ex] 2j = 240 \\[3ex] j = \dfrac{240}{2} \\[5ex] j = 120^\circ \\[3ex] (ii) \\[3ex] $Construction: Extend the two parallel lines in order to use angle properties between parallel lines$ h = \theta ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \theta + 120 = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \implies \\[3ex] h + 120 = 180 \\[3ex] h = 180 - 120 \\[3ex] h = 60^\circ \\[3ex] Also: \\[3ex] j = \phi ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \phi + k = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \implies \\[3ex] 120 + k = 180 \\[3ex] k = 180 - 120 \\[3ex] k = 60^\circ $(172.) NZQA Explain how we know from parts (i) and (ii) that the quadrilateral [from Question (171.)] must actually be a rhombus. You may wish to use the diagram below, which has the corners labelled. Based on Parts (i) and (ii) from Question (169.): Because the opposite angles of the quadrilateral are equal, the quadrilateral is likely a rhombus. (173.) NZQA Nails E, F, G, and H all lie on the circumference of a circle, with centre C. Angle FGE = 81° Angle GPF = 77° (i) Find the size, x, of angle HEG. Justify your answer. (ii) Find the size, y, of reflex angle ECF. Justify your answer.$ \angle FGP = \angle FGE = 81°...diagram \\[3ex] \angle GFP + \angle FGP + \angle GPF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FGP \\[3ex] \angle GFP + 81 + 77 = 180 \\[3ex] \angle GFP + 158 = 180 \\[3ex] \angle GFP = 180 - 158 \\[3ex] \angle GFP = 22^\circ \\[3ex] (i) \\[3ex] x = \angle GFP = 22^\circ ... \angle s\;\;in\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] Obtuse\;\;\angle FCE = 2 * \angle FGE ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle FCE = 2 * 81 \\[3ex] Obtuse\;\;\angle FCE = 162^\circ \\[3ex] y + Obtuse\;\;\angle FCE = 360^\circ ... \angle s\;\;at\;\;a\;\;point \\[3ex] y + 162 = 360 \\[3ex] y = 360 - 162 \\[3ex] y = 198^\circ $(174.) curriculum.gov.mt The points A, B, C and D lie on the circumference of a circle centre O AD and BC are two parallel lines. Angle$O\hat{A}D = 50^\circ$Calculate the values of angles a, b, c, d and e giving reasons for your answers and show your working where necessary.$ 1st: \\[3ex] a = 50^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] 2nd: \\[3ex] O\hat{D}A = O\hat{A}D = 50^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOD \\[3ex] O\hat{D}A + O\hat{A}D + A\hat{O}D = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] 50 + 50 + b = 180 \\[3ex] 100 + b = 180 \\[3ex] b = 180 - 100 \\[3ex] b = 80^\circ \\[3ex] 3rd: \\[3ex] b + c = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 80 + c = 180 \\[3ex] c = 180 - 80 \\[3ex] c = 100^\circ \\[3ex] OR \\[3ex] c = O\hat{D}A + O\hat{A}D \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] c = 50 + 50 \\[3ex] c = 100^\circ \\[3ex] 4th: \\[3ex] D\hat{O}C = A\hat{O}B ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \implies \\[3ex] D\hat{O}C = c \\[3ex] D\hat{O}C = 100^\circ \\[3ex] OR \\[3ex] D\hat{O}C = O\hat{D}A + O\hat{A}D \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] D\hat{O}C = 50 + 50 \\[3ex] D\hat{O}C = 100^\circ \\[3ex] O\hat{D}C = O\hat{C}D = p...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOD \\[3ex] O\hat{D}C + O\hat{C}D + D\hat{O}C = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] p + p + 100 = 180 \\[3ex] 2p = 180 - 100 \\[3ex] 2p = 80 \\[3ex] p = \dfrac{80}{2} \\[5ex] p = 40 \\[3ex] \therefore O\hat{D}C = O\hat{C}D = 40^\circ \\[3ex] \implies \\[3ex] O\hat{D}C = d = 40^\circ \\[3ex] 5th: \\[3ex] O\hat{C}D = e = 40^\circ $(175.) NZQA Nails Q, V, and W all lie on the circumference of a circle, with centre C Other nails form a straight line PQR which is tangent to the circle at point Q The lines PVW and QCW are straight. Angle QPV = 40° Find the size, e, of angle QCV Justify your answer with clear geometric reasoning$ \angle CQP = 90^\circ...radius\;CQ\;\;\perp\;\;tangent\;PQR\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \underline{\triangle PQW} \\[3ex] \angle QWP + \angle WQP + \angle WPQ = 180^\circ...sum\;\;of\;\;\angle s\;\;a\;\;\triangle \\[3ex] \angle WQP = 40^\circ ...diagram \\[3ex] \angle WQP = \angle CQP = 90^\circ...diagram \\[3ex] \implies \\[3ex] \angle QWP + 40 + 90 = 180 \\[3ex] \angle QWP + 130 = 180 \\[3ex] \angle QWP = 180 - 130 \\[3ex] \angle QWP = 50^\circ \\[3ex] \underline{\triangle CVW} \\[3ex] \angle CWV = \angle QWP = 50^\circ ... diagram \\[3ex] \angle CVW = \angle CWV = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\triangle \\[3ex] e = \angle CVW + \angle CWV ...exterior\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] e = 50 + 50 \\[3ex] e = 100^\circ $(176.) JAMB From the cyclic quadrilateral$TUVW$above, find the value of$x A.\:\: 24^\circ \\[3ex] B.\:\: 26^\circ \\[3ex] C.\:\: 20^\circ \\[3ex] D.\:\: 23^\circ \\[3ex]  \underline{Quadrilateral\:\:TUVW} \\[3ex] \angle U + \angle W = 180^\circ \\[3ex] ...opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:quad\;\;are\;\;supplementary \\[3ex] 3x + 20 + 88 = 180 \\[3ex] 3x + 108 = 180 \\[3ex] 3x = 180 - 108 \\[3ex] 3x = 72 \\[3ex] x = \dfrac{72}{3} \\[5ex] x = 24^\circ $(177.) SQA National 5 Maths AC is a tangent to the circle, centre O, with point of contact B. DE is a diameter of the circle and F is a point on the circumference. Angle ABD is 77° and angle DEF is 64° Calculate the size of angle BDF We can solve this question in at least two ways. Use any approach you prefer.$ \underline{First\;\;Method} \\[3ex] \angle ABD = \angle DEB = 77^\circ...\angle\;\;between\;\;tangent\;ABC\;\;and\;\;chord\;DB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle FEB = \angle DEF + \angle DEB ...diagram \\[3ex] \angle FEB = 64 + 77 \\[3ex] \angle FEB = 141^\circ \\[3ex] \angle BDF + \angle FEB = 180^\circ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BDF + 141 = 180 \\[3ex] \angle BDF = 180 - 141 \\[3ex] \angle BDF = 39^\circ \\[5ex] \underline{Second\;\;Method} \\[3ex] \angle DBE = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ABD + \angle DBE + \angle EBC = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 77 + 90 + \angle EBC = 180 \\[3ex] 167 + \angle EBC = 180 \\[3ex] \angle EBC = 180 - 167 \\[3ex] \angle EBC = 13^\circ \\[3ex] \angle EBC = \angle EDB = 13^\circ ... \angle\;\;between\;\;tangent\;ABC\;\;and\;\;chord\;BE = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DFE = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle EDF + \angle DFE + \angle DEF = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle\;DFE \\[3ex] \angle EDF + 90 + 64 = 180 \\[3ex] \angle EDF + 154 = 180 \\[3ex] \angle EDF = 180 - 154 \\[3ex] \angle EDF = 26^\circ \\[3ex] \angle BDF = \angle EDB + \angle EDF ... diagram \\[3ex] \angle BDF = 13 + 26 \\[3ex] \angle BDF = 39^\circ $(178.) curriculum.gov.mt This is a circle with centre O. What is the value of the angle y?$ 78 = 2 * y ...\angle\;\;at\;\;centre = 2 *\angle\;\;at\;\;circumference \\[3ex] 2y = 78 \\[3ex] y = \dfrac{78}{2} \\[5ex] y = 39^\circ $(179.) JAMB From the diagram above, find$x$O is the center of the circle UT is the tangent to the circle$ A.\;\; 55^\circ \\[3ex] B.\;\; 50^\circ \\[3ex] C.\;\; 75^\circ \\[3ex] D.\;\; 65^\circ \\[3ex] $We can solve this question in at least two ways. Use any approach you prefer. Construction is needed in either approach. First Method Construction: Join the radius from point T to center O Extend tangent UT to J$ \angle OTR = \angle ORT = 25^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OTJ = 90^\circ...radius\;OT \;\;\perp\;\;tangent\;UTJ\;\;at\;\;point\;\;of\;\;contant\;T \\[3ex] \angle OTJ = \angle OTR + x ...diagram \\[3ex] 90 = 25 + x \\[3ex] 25 + x = 90 \\[3ex] x = 90 - 25 \\[3ex] x = 65^\circ \\[3ex] $Second Method Construction: Join the chord from point T to point S$ \angle RTS = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle RST = x^\circ ... \angle\;\;between\;\;tangent\;UT\;\;and\;\;chord\;RT = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle RST + \angle RTS + \angle TRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RTS \\[3ex] x + 90 + 25 = 180 \\[3ex] x + 115 = 180 \\[3ex] x = 180 - 115 \\[3ex] x = 65^\circ $(180.) curriculum.gov.mt a) In the following diagram, calculate the values of angles p and q. Give reasons for your answers. b) In the following diagram, AB and CD are parallel. ∠BAP = 50° i) Write down the value of ∠BCD, giving a reason for your answer. ii) Explain why triangles ABP and PCD are both isosceles.$ a) \\[3ex] p + 78 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] p = 180 - 78 \\[3ex] p = 102^\circ \\[3ex] q = p = 102^\circ \\[3ex] ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] b) \\[3ex] i) \\[3ex] \angle BCD = 50^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] ii) \\[3ex] \angle ADC = 50^\circ ... alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] \angle PDC = \angle ADC = 50^\circ ... diagram \\[3ex] \angle PCD = \angle BCD = 50^\circ ... diagram \\[3ex] Because\;\; \angle PCD = 50^\circ \;\;and\;\; \angle PDC = 50^\circ \\[3ex] \triangle PCD \;\;is\;\;isosceles \\[3ex] Also: \\[3ex] \angle ABC = \angle ADC = 50^\circ ... alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] \angle ABP = \angle ABC = 50^\circ ... diagram \\[3ex] Because\;\; \angle ABP = 50^\circ \;\;and\;\; \angle BAP = 50^\circ \\[3ex] \triangle ABP \;\;is\;\;isosceles $(181.) WASSCE In the diagram, MNPQ is a circle with centre O |MN| = |NP| and ∠OMN = 50° Find: (i.) ∠MNP (ii.) ∠POQ$ \angle MPQ = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle QPN + \angle OMN = 180^\circ ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QPN + 50 = 180 \\[3ex] \angle QPN = 180 - 50 \\[3ex] \angle QPN = 130^\circ \\[3ex] \angle MPN + \angle MPQ = \angle QPN ...diagram \\[3ex] \angle MPN + 90 = 130 \\[3ex] \angle MPN = 130 - 90 \\[3ex] \angle MPN = 40^\circ \\[3ex] \angle MPN = \angle NMP = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] (i) \\[3ex] \angle MNP + \angle MPN + \angle NMP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNP \\[3ex] \angle MNP + 40 + 40 = 180 \\[3ex] \angle MNP + 80 = 180 \\[3ex] \angle MNP = 180 - 80 \\[3ex] \angle MNP = 100^\circ \\[5ex] \angle PQO + \angle MNP = 180^\circ ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle PQO + 100 = 180 \\[3ex] \angle PQO = 180 - 100 \\[3ex] \angle PQO = 80^\circ \\[3ex] \angle QPO = \angle PQO = 80^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] (ii) \\[3ex] \angle POQ + \angle QPO + \angle PQO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POQ \\[3ex] \angle POQ + 80 + 80 = 180 \\[3ex] \angle POQ + 160 = 180 \\[3ex] \angle POQ = 180 - 160 \\[3ex] \angle POQ = 20^\circ $(182.) GCSE P, Q and R are points on the circumference of the circle, centre O PO is parallel to QR and angle POQ = 48° (i) Find angle OPR (ii) The radius of the circle is 5.4 cm Calculate the length of the major arc PQ.$ (i) \\[3ex] \angle POQ = 2 * \angle PRQ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 48 = 2 * \angle PRQ \\[3ex] 2 * \angle PRQ = 48 \\[3ex] \angle PRQ = \dfrac{48}{2} \\[5ex] \angle PRQ = 24^\circ \\[3ex] \angle OPR = \angle PRQ...alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] \therefore \angle OPR = 24^\circ \\[3ex] (ii) \\[3ex] radius = r = 5.4\;cm \\[3ex] Acute\;\angle POQ + Reflex\;\angle POQ = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] 48 + Reflex\;\angle POQ = 360 \\[3ex] Reflex\;\angle POQ = 360 - 48 \\[3ex] Reflex\;\angle POQ = 312^\circ \\[3ex] Length\;\;of\;\;major\;\;arc\;PQ = \dfrac{Reflex\;\angle POQ}{360} * 2\pi * r \\[5ex] = \dfrac{312}{360} * 2 * \dfrac{22}{7} * 5.4 \\[5ex] = \dfrac{74131.2}{2520} \\[5ex] = 29.41714286\;cm $(183.) WASSCE The diagram shows a circle$O$and$\overline{PQ}$is a diameter. If$\angle PSO = 44^\circ$and |SR| = |RQ|, calculate the value of$\angle OQR$Construction: Join the chord from point S to point Q in order to use the isosceles triangle$ \underline{\triangle POS} \\[3ex] \angle PSO = \angle SPO = 44^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POS \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \angle P + \angle R = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle P = \angle SPO = 44^\circ \\[3ex] \implies \\[3ex] 44 + \angle R = 180 \\[3ex] \angle R = 180 - 44 \\[3ex] \angle R = 136^\circ \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ + \angle RQS + \angle SRQ = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] \angle RSQ = \angle RQS = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SRQ \\[3ex] \angle SRQ = \angle R = 136^\circ \\[3ex] \implies \\[3ex] k + k + 136 = 180 \\[3ex] 2k = 180 - 136 \\[3ex] 2k = 44 \\[3ex] k = \dfrac{44}{2} \\[5ex] k = 22 \\[3ex] \therefore \angle RSQ = \angle RQS = 22^\circ \\[3ex] \underline{\triangle SOQ} \\[3ex] \angle SOQ = \angle PSO + \angle SPO ...exterior\;\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle SOQ = 44 + 44 \\[3ex] \angle SOQ = 88^\circ \\[3ex] \angle OSQ + \angle OQS + \angle SOQ = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOQ \\[3ex] \angle OSQ = \angle OQS = m ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SOQ \\[3ex] \implies \\[3ex] m + m + 88 = 180 \\[3ex] 2m = 180 - 88 \\[3ex] 2m = 92 \\[3ex] m = \dfrac{92}{2} \\[5ex] m = 46 \\[3ex] \therefore \angle OSQ = \angle OQS = 46^\circ \\[3ex] \angle OQR = \angle OQS + \angle RQS ...diagram \\[3ex] \angle OQR = 46 + 22 \\[3ex] \angle OQR = 68^\circ $(184.) curriculum.gov.mt Work out the value of ∠STR.$ \underline{\triangle RPQ} \\[3ex] \angle RPQ = \angle RQP = k ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RPQ \\[3ex] \angle RPQ + \angle RQP + \angle PRQ = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle RPQ \\[3ex] k + k + 80 = 180 \\[3ex] 2k = 180 - 80 \\[3ex] 2k = 100 \\[3ex] k = \dfrac{100}{2} \\[5ex] k = 50 \\[3ex] \therefore \angle RPQ = \angle RQP = 50^\circ \\[3ex] \angle SPQ = \angle RPQ = 50^\circ ... diagram \\[3ex] \angle STR = \angle SPQ = 50^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $(185.) WASSCE In the diagram, TS is a tangent to the circle at S |PR| = |RS| and ∠PQR = 117° Calculate ∠PST$ A.\;\; 54^\circ \\[3ex] B.\;\; 44^\circ \\[3ex] C.\;\; 34^\circ \\[3ex] D.\;\; 27^\circ \\[3ex]  \angle PSR + 117 = 180^\circ ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle PSR = 180 - 117 \\[3ex] \angle PSR = 63^\circ \\[3ex] \angle SPR = \angle PSR = 63^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle PRS + \angle SPR + \angle PSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] \angle PRS + 63 + 63 = 180 \\[3ex] \angle PRS + 126 = 180 \\[3ex] \angle PRS = 180 - 126 \\[3ex] \angle PRS = 54^\circ \\[3ex] \angle PST = \angle PRS = 54^\circ ... \angle\;\;between\;\;tangent\;TS\;\;and\;\;chord\;PS = \angle\;\;in\;\;alternate\;\;segment $(186.) JAMB If$O$is the center of the circle in the figure above, find the value of$x A.\:\: 50^\circ \\[3ex] B.\:\: 260^\circ \\[3ex] C.\:\: 100^\circ \\[3ex] D.\:\: 65^\circ \\[3ex] E.\:\: 130^\circ \\[3ex]  y = 2(130) ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] y = 260^\circ \\[3ex] x + y = 360 ...\angle s \:\:at\:\:a\:\:point \\[3ex] x = 360 - y \\[3ex] x = 360 - 260 \\[3ex] x = 100^\circ $(187.) SQA National 5 Maths In the diagram shown below: ABE is a tangent to the circle centre O Angle DBE is 58° Calculate the size of angle CAB.$ \angle DCB = 58^\circ ...\angle\;\;between\;\;tangent\;ABE\;\;and\;\;chord\;BD = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle OCB = \angle DCB = 58^\circ ...diagram \\[3ex] \angle OBC = \angle OCB = 58^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle\;OCB \\[3ex] \angle OBA = 90^\circ ...radius\;OB\;\;\perp\;\;tangent\;ABE\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] But: \\[3ex] \angle OBA = \angle OBC + \angle CBA ...diagram \\[3ex] 90 = 58 + \angle CBA \\[3ex] \angle CBA + 58 = 90 \\[3ex] \angle CBA = 90 - 58 \\[3ex] \angle CBA = 32^\circ \\[3ex] \angle OCB = \angle CAB + \angle CBA ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] 58 = \angle CAB + 32 \\[3ex] \angle CAB + 32 = 58 \\[3ex] \angle CAB = 58 - 32 \\[3ex] \angle CAB = 26^\circ $(188.) ACT A circle, 2 chords, and some lengths, in centimeters, are shown in the figure below, which is not drawn to scale. What is the value of x? (Note: When two chords intersect, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.)$ A.\;\; 10 \\[3ex] B.\;\; 13.5 \\[3ex] C.\;\; 14 \\[3ex] D.\;\; 17.5 \\[3ex] E.\;\; 19 \\[3ex]  2 * x = 7 * 5 ...Intersecting\;\;Chords\;\;Theorem \\[3ex] x = \dfrac{7 * 5}{2} \\[5ex] x = \dfrac{35}{2} \\[5ex] x = 17.5 $(189.) JAMB P, R, and S lie on a circle centre O as shown above, while Q lies outside the circle. Find ∠PSO$ A.\;\; 40^\circ \\[3ex] B.\;\; 45^\circ \\[3ex] C.\;\; 55^\circ \\[3ex] D.\;\; 35^\circ \\[3ex]  \angle PRS = 20 + 35 ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle PRS = 55^\circ \\[3ex] \angle POS = 2 * \angle PRS ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle POS = 2 * 55 \\[3ex] \angle POS = 110^\circ \\[3ex] \angle PSO = \angle SPO = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle PSO + \angle SPO + \angle POS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POS \\[3ex] k + k + 110 = 180 \\[3ex] 2k = 180 - 110 \\[3ex] 2k = 70 \\[3ex] k = \dfrac{70}{2} \\[5ex] k = 35 \\[3ex] \therefore \angle PSO = 35^\circ $(190.) WASSCE The diagram is a circle with centre O If ∠SPR = 2m and ∠SQR = n, express m in terms of n$ A.\;\; m = \dfrac{n}{2} \\[5ex] B.\;\; m = 2n \\[3ex] C.\;\; m = n - 2 \\[3ex] D.\;\; m = n + 2 \\[3ex]  2m = n ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] m = \dfrac{n}{2} $(191.) JAMB Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31° and ∠PKR = 42°, find ∠KQR$ A.\;\; 11^\circ \\[3ex] B.\;\; 73^\circ \\[3ex] C.\;\; 107^\circ \\[3ex] D.\;\; 138^\circ \\[3ex] E.\;\; 149^\circ \\[3ex] $It is very necessary to draw the diagram$ \angle KRQ = 31^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment \\[3ex] \angle RKQ + 42 = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RKQ = 180 - 42 \\[3ex] \angle RKQ = 138^\circ \\[3ex] \angle KQR + \angle RKQ + \angle KRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RKQ \\[3ex] \angle KQR + 31 + 138 = 180 \\[3ex] \angle KQR + 169 = 180 \\[3ex] \angle KQR = 180 - 169 \\[3ex] \angle KQR = 11^\circ $(192.) WASSCE In the diagram,$\overline{AB}$is a tangent to the circle with centre O and COB is a straight line. If CD || AB and ∠ABE = 40°, find ∠ODE$ \angle ECD = \angle ABE = 40^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OCD = \angle ECD = 40^\circ ...diagram \\[3ex] \angle ODC = \angle OCD = 40^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle CDE = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CDE = \angle ODC + \angle ODE ...diagram \\[3ex] 90 = 40 + \angle ODE \\[3ex] \angle ODE + 40 = 90 \\[3ex] \angle ODE = 90 - 40 \\[3ex] \angle ODE = 50^\circ $(193.) curriculum.gov.mt In the diagram below, points P, Q, R and S lie on a circle centre O. PS is a diameter and RT is a tangent to the circle. Lines RS and QO are parallel.$R\hat{S}P = 34^\circ$Work out the value of the angles marked a, b, c and d, giving reasons.$ \angle PRS = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] a + \angle PRS + \angle PSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] a + 90 + 34 = 180 \\[3ex] a + 124 = 180 \\[3ex] a = 180 - 124 \\[3ex] a = 56^\circ \\[5ex] \angle QOP = \angle RSP = 34^\circ.. corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] b + \angle QOP = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] b + 34 = 180 \\[3ex] b = 180 - 34 \\[3ex] b = 146^\circ \\[3ex] c + 34 = 180^\circ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] c = 180 - 34 \\[3ex] c = 146^\circ \\[3ex] d = a = 56^\circ ...\angle\;\;between\;\;tangent\;RT\;\;and\;\;chord\;RS = \angle\;\;in\;\;alternate\;\;segment $(194.) JAMB In the diagram above, O is the centre of the circle ∠UOT = 70° and ∠RST = 100° Calculate ∠RUO$ A.\;\; 20^\circ \\[3ex] B.\;\; 25^\circ \\[3ex] C.\;\; 50^\circ \\[3ex] D.\;\; 80^\circ \\[3ex]  \angle OUT = \angle OTU = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OUT + \angle OTU + \angle UOT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] p + p + 70 = 180 \\[3ex] 2p = 180 - 70 \\[3ex] 2p = 110 \\[3ex] p = \dfrac{110}{2} \\[5ex] p = 55^\circ \\[3ex] \angle RUT + \angle RST = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle RUT + 100 = 180 \\[3ex] \angle RUT = 180 - 100 \\[3ex] \angle RUT = 80^\circ \\[3ex] \angle RUT = \angle RUO + \angle OUT ...diagram \\[3ex] 80 = \angle RUO + 55 \\[3ex] \angle RUO + 55 = 80 \\[3ex] \angle RUO = 80 - 55 \\[3ex] \angle RUO = 25^\circ $(195.) WASSCE In the diagram, O is the centre of the circle PQR ∠RPQ = 73° and ∠POQ = 62° Calculate ∠RQO$ A.\;\; 11^\circ \\[3ex] B.\;\; 14^\circ \\[3ex] C.\;\; 17^\circ \\[3ex] D.\;\; 31^\circ \\[3ex]  \angle POQ = 2 * \angle PRQ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 62 = 2 * \angle PRQ \\[3ex] 2 * \angle PRQ = 62 \\[3ex] \angle PRQ = \dfrac{62}{2} \\[5ex] \angle PRQ = 31^\circ \\[3ex] \angle PQR + \angle RPQ + \angle PRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRQ \\[3ex] \angle PQR + 73 + 31 = 180 \\[3ex] \angle PQR + 104 = 180 \\[3ex] \angle PQR = 180 - 104 \\[3ex] \angle PQR = 76^\circ \\[3ex] \angle PQO = \angle QOP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OPQ \\[3ex] \angle PQO + \angle QOP + \angle POQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPQ \\[3ex] k + k + 62 = 180 \\[3ex] 2k = 180 - 62 \\[3ex] 2k = 118 \\[3ex] k = \dfrac{118}{2} \\[5ex] k = 59 \\[3ex] \therefore \angle PQO = 59^\circ \\[3ex] \angle PQR = \angle PQO + \angle RQO ...diagram \\[3ex] 76 = 59 + \angle RQO \\[3ex] 59 + \angle RQO = 76 \\[3ex] \angle RQO = 76 - 59 \\[3ex] \angle RQO = 17^\circ $(196.) JAMB In the diagram above, PQR is a straight line and$\overline{PS}$is a tangent to the circle QRS with |PS| = |SR| and ∠SPR = 40° Find ∠PSQ$ A.\;\; 20^\circ \\[3ex] B.\;\; 40^\circ \\[3ex] C.\;\; 10^\circ \\[3ex] D.\;\; 30^\circ \\[3ex]  \underline{First\;\;Approach} \\[3ex] \angle SRP = \angle SPR = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle PSR + \angle SPR + \angle SRP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] \angle PSR + 40 + 40 = 180 \\[3ex] \angle PSR + 80 = 180 \\[3ex] \angle PSR = 180 - 80 \\[3ex] \angle PSR = 100^\circ \\[3ex] \angle QSR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PSR = \angle PSQ + \angle QSR ...diagram \\[3ex] 100 = \angle PSQ + 90 \\[3ex] \angle PSQ + 90 = 100 \\[3ex] \angle PSQ = 100 - 90 \\[3ex] \angle PSQ = 10^\circ \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle SPQ = \angle SPR = 40^\circ ... diagram \\[3ex] \angle SRP = \angle SPR = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle SPQ = \angle SPR = 40^\circ \\[3ex] \angle QSR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PQS = \angle QSR + \angle QRS \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle PQS = 90 + 40 \\[3ex] \angle PQS = 130^\circ \\[3ex] \angle PSQ + \angle PQS + \angle SPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] \angle PSQ + 130 + 40 = 180 \\[3ex] \angle PSQ + 170 = 180 \\[3ex] \angle PSQ = 180 - 170 \\[3ex] \angle PSQ = 10^\circ $(197.) WASSCE In the figure, O is the centre of the circle SRP, ∠ROS = 110° and ∠QRP is a right angle. Find the value of x$ A.\;\; 70 \\[3ex] B.\;\; 55 \\[3ex] C.\;\; 40 \\[3ex] D.\;\; 35 \\[3ex] $We can solve this question using at least two approaches. Choose any approach you prefer.$ \underline{First\;\;Approach} \\[3ex] \angle ROS = 2 * \angle RPS ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 110 = 2 * \angle RPS \\[3ex] 2 * \angle RPS = 110 \\[3ex] \angle RPS = \dfrac{110}{2} \\[5ex] \angle RPS = 55^\circ \\[3ex] \angle RPQ = \angle RPS = 55^\circ ...diagram \\[3ex] \angle PQR + \angle QRP + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] x + 90 + 55 = 180 \\[3ex] x + 145 = 180 \\[3ex] x = 180 - 145 \\[3ex] x = 35^\circ \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle ROS + \angle POS = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 110 + \angle POS = 180 \\[3ex] \angle POS = 180 - 110 \\[3ex] \angle POS = 70^\circ \\[3ex] \angle OPS = \angle OSP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OPS \\[3ex] \angle OPS + \angle OSP + \angle POS = 180^\circ ..sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPS \\[3ex] k + k + 70 = 180 \\[3ex] 2k = 180 - 70 \\[3ex] 2k = 110 \\[3ex] k = \dfrac{110}{2} \\[5ex] k = 55 \\[3ex] \therefore \angle OPS = 55^\circ \\[3ex] \angle RPQ = \angle OPS = 55^\circ ...diagram \\[3ex] \angle PQR + \angle QRP + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] x + 90 + 55 = 180 \\[3ex] x + 145 = 180 \\[3ex] x = 180 - 145 \\[3ex] x = 35^\circ $(198.) JAMB In the diagram above, PQR is a circle with centre O If$\angle QPR\;\;is\;\; x^\circ$, find$\angle QRP A.\;\; x^\circ \\[3ex] B.\;\; (90 - x)^\circ \\[3ex] C.\;\; (90 + x)^\circ \\[3ex] D.\;\; (180 - x)^\circ \\[3ex]  \angle PQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle QRP + \angle QPR + \angle PQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle QRP + x + 90 = 180 \\[3ex] \angle QRP = 180 - x - 90 \\[3ex] \angle QRP = 90 - x \\[3ex] \angle QRP = (90 - x)^\circ $(199.) SQA National 5 Maths Pam is designing a company logo. She starts by drawing a regular pentagon ABCDE. The vertices of the pentagon lie on the circumference of a circle with centre O. She then adds to the design as shown in the diagram below. AF is a diameter of the circle. Calculate the size of angle OFB.$ Sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;regular\;\;Pentagon\;\;ABCDE \\[3ex] = 180(5 - 2) \\[3ex] = 180(3) \\[3ex] = 540^\circ \\[3ex] Each\;\;\angle \\[3ex] = \dfrac{540}{5} \\[5ex] = 108^\circ \\[3ex] \implies \\[3ex] \angle A = 108^\circ \\[3ex] \underline{Second\;\;Diagram} \\[3ex] Reflex\;\;\angle EOB = 2 * \angle A ... \angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle EOB = 2(108) = 216^\circ \\[3ex] Acute\;\;\angle EOB + Reflex\;\;\angle EOB = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Acute\;\;\angle EOB + 216 = 360^\circ \\[3ex] Acute\;\;\angle EOB = 360 - 216 = 144 \\[3ex] \angle OFE = \angle OEF = \angle OFB = \angle OBF = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle s\;\;OEF\;\;and\;\;OBF \\[3ex] \angle EFB = \angle OFE + \angle OFB ...diagram \\[3ex] Acute\;\;\angle EOB = 2 * \angle EFB ... \angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 144 = 2 * (p + p) \\[3ex] 144 = 2(2p) \\[3ex] 144 = 4p \\[3ex] 4p = 144 \\[3ex] p = \dfrac{144}{4} \\[5ex] p = 36 \\[3ex] \therefore \angle OFB = 36^\circ $(200.) ACT Right triangle$\triangle ABC$is inscribed in a circle with center M, as shown below, and C can be any point on the circle other than A or B Which of the following is the most direct explanation of why$\triangle MCA$is isosceles? F. 2 sides are radii of the circle G. Side-angle-side congruence H. Angle-side-angle congruence J. Angle-angle-angle similarity K. The Pythagorean theorem Construction: Draw the radius from center: M to circumference: C Because MC and MA are radii (same length):$\angle MCA = \angle MAC$...base angles of isosceles$\triangle MCA$(201.) WASSCE In the diagram, PS is tangent to the circle of centre O If QS is a straight line and ∠TOR = 82°, find ∠RST We can solve this question in at least two ways. Use any approach you prefer$ \underline{First\;\;Approach} \\[3ex] \angle TOR = 2 * \angle TQR ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 82 = 2 * \angle TQR \\[3ex] 2 * \angle TQR = 82 \\[3ex] \angle TQR = \dfrac{82}{2} \\[5ex] \angle TQR = 41^\circ \\[3ex] \angle TQS = \angle TQR = 41^\circ ...diagram \\[3ex] \angle OTS = 90^\circ...radius\;OT \perp \;\;tangent\;PTS\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle QTS = \angle OTS = 90^\circ ... diagram \\[3ex] \angle QST + \angle QTS + \angle TQS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QTS \\[3ex] \angle QST + 90 + 41 = 180 \\[3ex] \angle QST + 131 = 180 \\[3ex] \angle QST = 180 - 131 \\[3ex] \angle QST = 49^\circ \\[3ex] \angle RST = \angle QST = 49^\circ \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle OTS = 90^\circ...radius\;OT \perp \;\;tangent\;PTS\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle OTR = \angle ORT = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OTR + \angle ORT + \angle TOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TOR \\[3ex] p + p + 82 = 180 \\[3ex] 2p = 180 - 82 \\[3ex] 2p = 98 \\[3ex] p = \dfrac{98}{2} \\[5ex] p = 49 \\[3ex] \therefore \angle OTR = 49^\circ \\[3ex] \angle OTS = \angle OTR + \angle RTS ...diagram \\[3ex] 90 = 49 + \angle RTS \\[3ex] 49 + \angle RTS = 90 \\[3ex] \angle RTS = 90 - 49 \\[3ex] \angle RTS = 41^\circ \\[3ex] \angle QRT = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle QRT + \angle TRS = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + \angle TRS = 180 \\[3ex] \angle TRS = 180 - 90 \\[3ex] \angle TRS = 90^\circ \\[3ex] \angle RTS + \angle TRS + \angle RST = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TRS \\[3ex] 41 + 90 + \angle RST = 180 \\[3ex] 131 + \angle RST = 180 \\[3ex] \angle RST = 180 - 131 \\[3ex] \angle RST = 49^\circ $(202.) WASSCE In the diagram, O is the centre of the circle ABCDE$|\overline{BC}| = |\overline{CD}|$and ∠BCD = 108° Find ∠CDE Construction: Join the chord from point B to point D$ \angle CBD = \angle CDB = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle CBD + \angle CDB + \angle BCD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BCD \\[3ex] k + k + 108 = 180 \\[3ex] 2k = 180 - 108 \\[3ex] 2k = 72 \\[3ex] k = \dfrac{72}{2} \\[5ex] k = 36 \\[3ex] \therefore \angle CDB = 36^\circ \\[3ex] \angle BDE = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CDE = \angle CDB + \angle BDE ...diagram \\[3ex] \angle CDE = 36 + 90 \\[3ex] \angle CDE = 126^\circ $(203.) WASSCE The diagram shows a circle ABCD with centre E Quadrilateral EADC is a rhombus, ∠BAE = ∠ECB = n and ∠ABC = m Find: (i.) m (ii.) n$ (i) \\[3ex] Obtuse\;\angle AEC = 2 * m = 2m...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\angle AEC + Obtuse\;\angle AEC = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\angle AEC + 2m = 360 \\[3ex] Reflex\;\angle AEC = 360 - 2m \\[5ex] Reflex\;\angle AEC = 2 * \angle ADC ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 360 - 2m = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 360 - 2m \\[3ex] \angle ADC = \dfrac{360 - 2m}{2} \\[5ex] \angle ADC = \dfrac{2(180 - m)}{2} \\[5ex] \angle ADC = 180 - m \\[5ex] Obtuse\;\angle AEC = \angle ADC ... opposite\;\;\angle s\;\;of\;\;Rhombus\;EADC\;\;are\;\;equal \\[3ex] \implies \\[3ex] 2m = 180 - m \\[3ex] 2m + m = 180 \\[3ex] 3m = 180 \\[3ex] m = \dfrac{180}{3} \\[3ex] m = 60^\circ \\[3ex] (ii) \\[3ex] \angle EAC = \angle ECA = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle EAC \\[3ex] \angle EAC + \angle ECA + \angle AEC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EAC \\[3ex] k + k + 2m = 180 \\[3ex] 2k = 180 - 2m \\[3ex] k = \dfrac{180 - 2m}{2} \\[5ex] k = \dfrac{2(90 - m)}{2} \\[5ex] k = 90 - m \\[3ex] \therefore \angle EAC = \angle ECA = 90 - m \\[3ex] \underline{\triangle ABC} \\[3ex] \angle EAC + n + \angle ECA + n + m = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] (90 - m) + n + (90 - m) + n + m = 180 \\[3ex] 90 - m + 90 - m + m + 2n = 180 \\[3ex] 2n - m + 180 = 180 \\[3ex] 2n - m = 180 - 180 \\[3ex] 2n - m = 0 \\[3ex] 2n = m \\[3ex] n = \dfrac{m}{2} \\[5ex] n = \dfrac{60}{2} \\[5ex] n = 30^\circ $(204.) JAMB PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS?$ A.\;\; 180^\circ \\[3ex] B.\;\; 150^\circ \\[3ex] C.\;\; 105^\circ \\[3ex] D.\;\; 75^\circ \\[3ex] $It is necessary to draw the diagram$ \angle QRS + \angle QPS = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QRS + 75 = 180 \\[3ex] \angle QRS = 180 - 75 \\[3ex] \angle QRS = 105^\circ $(205.) WASSCE In the diagram,$\overline{YW}$is a tangent to the circle at X, |UV| = |VX| and ∠VXW = 50° Find the value ∠UXY$ A.\;\; 70^\circ \\[3ex] B.\;\; 80^\circ \\[3ex] C.\;\; 105^\circ \\[3ex] D.\;\; 110^\circ \\[3ex]  \angle XUV = 50^\circ... \angle\;\;between\;\;tangent\;YXW\;\;and\;\;chord\;XV = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle UXV = \angle XUV = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle UXY + \angle UXV + 50 = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle UXY + 50 + 50 = 180 \\[3ex] \angle UXY + 100 = 180 \\[3ex] \angle UXY = 180 - 100 \\[3ex] \angle UXY = 80^\circ $(206.) WASSCE In the diagram, O is the centre of the circle, ∠XOZ = (10m)° and ∠XWZ = m° Calculate the value of m$ A.\;\; 30 \\[3ex] B.\;\; 36 \\[3ex] C.\;\; 40 \\[3ex] D.\;\; 72 \\[3ex]  Reflex\;\;\angle XOZ = 2 * m \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle XOZ = 2m \\[3ex] Reflex\;\;\angle XOZ + Obtuse\;\;\angle XOZ = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] 2m + 10m = 360 \\[3ex] 12m = 360 \\[3ex] m = \dfrac{360}{12} \\[5ex] m = 30 $(207.) WASSCE In the diagram, PQ is a tangent to the circle at Z If PQ || WY, ∠WZY = 120°, ∠WXY = m, &WYZ = y and ∠YZQ = x, find the value of: (a.) m (b.) x$ (a.) \\[3ex] m + 120 = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] m = 180 - 120 \\[3ex] m = 60^\circ \\[5ex] (b.) \\[3ex] x = \angle YWZ ...\angle\;\;between\;\;tangent\;PQ\;\;and\;\;chord\;YZ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] y = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle YWZ + \angle WYZ + \angle WZY = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WYZ \\[3ex] x + y + 120 = 180 \\[3ex] x + x = 180 - 120 \\[3ex] 2x = 60 \\[3ex] x = \dfrac{60}{2} \\[5ex] x = 30^\circ $(208.) NYSED Regents Examination In the diagram below, chords$\overline{PQ}$and$\overline{RS}$of circle O intersect at T. Which relationship must always be true?$ (1)\;\; RT = TQ \hspace{7em} (3)\;\; RT + TS = PT + TQ \\[3ex] (2)\;\; RT = TS \hspace{7em} (4)\;\; RT * TS = PT * TQ \\[3ex]  RT * TS = PT * TQ ...Intersecting\;\;Chords\;\;Theorem $(209.) CSEC The diagram below shows a circle, with the points P, Q, R and S lying on its circumference and its centre marked O RP is a diameter of the circle and AB is a tangent to the circle at P Angle APQ = 3x°, angle QPR = 2x°, angle RPS = x° and angle QSP = 54° Determine the value of EACH of the following angles. Show detailed working where possible and give a reason for your answer. (i.) x (ii.) y (iii.) z$ (i.) \\[3ex] \angle RSP = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle RSP = \angle RSQ + \angle QSP ...diagram \\[3ex] \implies \\[3ex] 90 = \angle QSR + 54 \\[3ex] \angle QSR + 54 = 90 \\[3ex] \angle QSR = 90 - 54 \\[3ex] \angle QSR = 36^\circ \\[3ex] \angle QPR = \angle QSR = 36^\circ ... \angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \implies \\[3ex] 2x = 36 \\[3ex] x = \dfrac{36}{2} \\[5ex] x = 18^\circ \\[3ex] (ii.) \\[3ex] \angle SRP + \angle RSP + \angle RPS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRP \\[3ex] y + 90 + x = 180 \\[3ex] y + 90 + 18 = 180 \\[3ex] y + 108 = 180 \\[3ex] y = 180 - 108 \\[3ex] y = 72^\circ \\[3ex] (iii.) \\[3ex] \angle SQP = \angle SRP ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \implies \\[3ex] z = y \\[3ex] z = 72^\circ $(210.) WASSCE In diagram, O is the centre of the circle, ∠QPS = 100°, ∠PSQ = 60° and ∠QSR = 80° Calculate ∠SQR$ A.\;\; 20^\circ \\[3ex] B.\;\; 40^\circ \\[3ex] C.\;\; 60^\circ \\[3ex] D.\;\; 80^\circ \\[3ex]  \angle QRS + \angle QPS = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QRS + 100 = 180 \\[3ex] \angle QRS = 180 - 100 \\[3ex] \angle QRS = 80^\circ \\[3ex] \angle SQR + \angle QSR + \angle QRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QRS \\[3ex] \angle SQR + 80 + 80 = 180 \\[3ex] \angle SQR + 160 = 180 \\[3ex] \angle SQR = 180 - 160 \\[3ex] \angle SQR = 20^\circ $(211.) NYSED Regents Examination In the diagram below of circle K, secand$\overline{PLKE}$and tangent$\overline{PZ}$are drawn from external point P. If$m\overset{\huge\frown}{LZ} = 56^\circ$, determine and state the degree measure of angle P. Construction: Join radius$\overline{KZ}$from center K to point Z$ m\overset{\huge\frown}{LZ} = \angle PKZ = 56^\circ \\[3ex] \angle KZP = 90^\circ ...radius\;KZ \perp tangent\;PZ\;\;at\;\;point\;\;of\;\;contact\;Z \\[3ex] \underline{\triangle KZP} \\[3ex] \angle KPZ + \angle PKZ + \angle KZP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle KPZ + 56 + 90 = 180 \\[3ex] \angle KPZ + 146 = 180 \\[3ex] \angle KPZ = 180 - 146 \\[3ex] \angle KPZ = 34 \\[3ex] \angle P = \angle KPZ = 34^\circ ... diagram $(212.) curriculum.gov.mt A quadrilateral is inscribed in a circle. Work out the value of x.$ 4x + 2x = 180^\circ \\[3ex] ...interior\:\:opposite\;\;\angle s\:\:of\:\:a\:\:cyclic\:\:quad\;\;are\;\;supplementary \\[3ex] 6x = 180 \\[3ex] x = \dfrac{180}{6} \\[5ex] x = 30^\circ $(213.) GCSE A and B are points on a circle. PA and PB are tangents. Work out the size of angle APB.$ PA = PB ...two\;\;tangents:\;\;PA\;\;and\;\;PB\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \angle PBA = \angle PAB = 71^\circ ...base \angle s\;\;of\;\;isosceles\;\triangle PAB \\[3ex] \angle APB + \angle PBA + \angle PAB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PAB \\[3ex] \angle APB + 71 + 71 = 180 \\[3ex] \angle APB + 142 = 180 \\[3ex] \angle APB = 180 - 142 \\[3ex] \angle APB = 38^\circ $(214.) WASSCE The diagram shows a triangle inscribed in a circle of centre O If$\angle LNM = (x + 15)^\circ$and$\angle NLM = (x - 5)^\circ$, find the value of$x A.\;\; 35^\circ \\[3ex] B.\;\; 40^\circ \\[3ex] C.\;\; 45^\circ \\[3ex] D.\;\; 50^\circ \\[3ex]  \angle LMN = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle LNM + \angle NLM + \angle LMN = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle LMN \\[3ex] (x + 15) + (x - 5) + 90 = 180 \\[3ex] x + 15 + x - 5 + 90 = 180 \\[3ex] 2x + 100 = 180 \\[3ex] 2x = 180 - 100 \\[3ex] 2x = 80 \\[3ex] x = \dfrac{80}{2} \\[5ex] x = 40^\circ $(215.) NYSED Regents Examination In circle O two secants,$\overline{ABP}$and$\overline{CDP}$, are drawn to external point P. If m$\overset{\huge\frown}{AC} = 72^\circ$, and m$\overset{\huge\frown}{BD} = 34^\circ$, what is the measure of ∠P?$ (1)\;\; 19^\circ \hspace{7em} (3)\;\; 53^\circ \\[3ex] (2)\;\; 38^\circ \hspace{7em} (4)\;\; 106^\circ \\[3ex] $Let us draw the circle O is the center of the circle$ \underline{Central\;\;\angle = Intercepted\;\; \overset{\huge\frown}{}} \\[3ex] \overset{\huge\frown}{AC} = \angle AOC = 72^\circ \\[3ex] \overset{\huge\frown}{BD} = \angle BOD = 34^\circ \\[3ex] \underline{\angle\;\;of\;\;Intersecting\;\;Secants\;(Outside\;\;the\;\;Circle)} \\[3ex] \angle P = \dfrac{72 - 34}{2} \\[5ex] \angle P = \dfrac{38}{2} \\[5ex] \angle P = 19^\circ $(216.) curriculum.gov.mt A quadrilateral is inscribed in a circle. Write down an expression for y in terms of x.$ 2y + 3x = 180^\circ \\[3ex] ...interior\:\:opposite\;\;\angle s\:\:of\:\:a\:\:cyclic\:\:quad\;\;are\;\;supplementary \\[3ex] 2y = 180 - 3x \\[3ex] 2y = 3(60 - x) \\[3ex] y = \dfrac{3(60 - x)}{2} $(217.) GCSE C, D and E are points on a different circle. Is X the centre of the circle? Tick a box. Show working to support your answer.$ \underline{1st\;\;Case} \\[3ex] \angle X + 204 = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] \angle X = 360 - 204 \\[3ex] \angle X = 156^\circ \\[3ex] \underline{2nd\;\;Case} \\[3ex] \angle X = 2(78^\circ) ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle X = 156^\circ \\[3ex] Because\;\;\angle X\;\;in\;\;2nd\;\;Case = \angle X\;\;in\;\;1st\;\;Case: \\[3ex] \angle X \;\;is\;\;the\;\;centre\;\;of\;\;the\;\;circle $(218.) WASSCE In the figure, PQRS is a cyclic quadrilateral, |QR| = |SR|, ∠SPT = x° and ∠QSR = 37° Find the value of x$ A.\;\; 143 \\[3ex] B.\;\; 106 \\[3ex] C.\;\; 90 \\[3ex] D.\;\; 74 \\[3ex]  \angle SQR = \angle QSR = 37^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RSQ \\[3ex] \angle SRQ + \angle SQR + \angle QSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RSQ \\[3ex] \angle SRQ + 37 + 37 = 180 \\[3ex] \angle SRQ + 74 = 180 \\[3ex] \angle SRQ = 180 - 74 \\[3ex] \angle SRQ = 106^\circ \\[3ex] x = 106^\circ \\[3ex] ... exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quad = interior\;\;opposite\;\;\angle $(219.) WASSCE In the diagram, RP is a diameter of the circle RSP, RP is produced to T and TS is a tangent to the circle at S If ∠PRS = 24°, calculate the value of ∠STR$ A.\;\; 24^\circ \\[3ex] B.\;\; 42^\circ \\[3ex] C.\;\; 48^\circ \\[3ex] D.\;\; 66^\circ \\[3ex] $Construction: Draw the chord from point S to point P$ \angle RSP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PST = 24^\circ ...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SP = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle RST = \angle RSP + \angle PST ... diagram \\[3ex] \angle RST = 90 + 24 \\[3ex] \angle RST = 114^\circ \\[3ex] \angle TRS = \angle PRS = 24^\circ ... diagram \\[3ex] \angle STR + \angle RST + \angle TRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RST \\[3ex] \angle STR + 114 + 24 = 180 \\[3ex] \angle STR + 138 = 180 \\[3ex] \angle STR = 180 - 138 \\[3ex] \angle STR = 42^\circ $(220.) NSC In the diagram, MP is a diameter of a circle centered at O MP cuts the chord NR at T Radius NO and chords PR, MN and MR are drawn.$\hat{R_1} = 69^\circ$Determine, giving reasons, the size of: (220.1.1)$\hat{R_2}$(220.1.2)$\hat{O_1}$(220.1.3)$\hat{M_1}$(220.1.4)$\hat{M_2}$, if it is further given that NO || PR$ (220.1.1) \\[3ex] \hat{R_1} + \hat{R_2} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] 69 + \hat{R_2} = 90 \\[3ex] \hat{R_2} = 90 - 69 \\[3ex] \hat{R_2} = 21^\circ \\[3ex] (220.1.2) \\[3ex] \hat{O_1} = 2 * \hat{R_1} ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \hat{O_1} = 2 * 69 \\[3ex] \hat{O_1} = 138^\circ \\[3ex] OR \\[3ex] \hat{M_1} = \hat{R_2} = 21^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \hat{N_1} = \hat{M_1} = 21^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OMN \\[3ex] \hat{O_1} + \hat{M_1} + \hat{N_1} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OMN \\[3ex] \hat{O_1} + 21 + 21 = 180 \\[3ex] \hat{O_1} + 42 = 180 \\[3ex] \hat{O_1} = 180 - 42 \\[3ex] \hat{O_1} = 138^\circ \\[3ex] (220.1.3) \\[3ex] \hat{M_1} = \hat{R_2} = 21^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (220.1.4) \\[3ex] \hat{O_2} + \hat{O_1} = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \hat{O_2} + 138 = 180 \\[3ex] \hat{O_2} = 180 - 138 \\[3ex] \hat{O_2} = 42^\circ \\[3ex] \angle MPR = \hat{O_2} = 42^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle PMR + \angle MRP + \angle MPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MRP \\[3ex] \hat{M_2} + (\hat{R_1} + \hat{R_2}) + 42 = 180 \\[3ex] \hat{M_2} + 90 + 42 = 180 \\[3ex] \hat{M_2} + 132 = 180 \\[3ex] \hat{M_2} = 180 - 132 \\[3ex] \hat{M_2} = 48^\circ $(221.) NYSED Regents Examination As shown in the diagram below, secants$\overrightarrow{PWR}$and$\overrightarrow{PTS}$are drawn to circle O from external point P. If$m\angle RPS = 35^\circ$and$m\overset{\huge\frown}{RS} = 121^\circ$, determine and state$m\overset{\huge\frown}{WT} Major\;\;arc = \overset{\huge\frown}{RS} = 121^\circ \\[3ex] Minor\;\;arc = \overset{\huge\frown}{WT} = ? \\[3ex] \underline{\angle\;\;of\;\;Intersecting\;\;Secants\;(Outside\;\;the\;\;Circle)} \\[3ex] \angle RPS = \dfrac{Major\;\;arc - Minor\;\;arc}{2} \\[5ex] 35 = \dfrac{121 - \overset{\huge\frown}{WT}}{2} \\[5ex] 35(2) = 121 - \overset{\huge\frown}{WT} \\[3ex] 70 = 121 - \overset{\huge\frown}{WT} \\[3ex] \overset{\huge\frown}{WT} = 121 - 70 \\[3ex] \overset{\huge\frown}{WT} = 51 \\[3ex] m\overset{\huge\frown}{WT} = 51^\circ $(222.) GCSE (a) Points A, B, C, P, Q and R lie on the circumference of the circle, centre O. Show that triangle ABC is congruent to triangle PQR. Give a reason for each step of your answer. (b) William says, "It is possible to draw one circle through the four vertices of any kite that has two opposite angles that are right angles." Is William correct? Show how you decide. (a) There are at least three ways to show congruency Use any approach you prefer$ \underline{First\;\;Approach:\;\;HL} \\[3ex] \angle ABC = \angle PQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \therefore \triangle ABC \;\;and\;\;\triangle PQR \;\;are\;\;right\;\;\triangle s \\[3ex] \implies \\[3ex] |AC| = |PR| = diameter = hypotenuse \;(H) \\[3ex] |AB| = |PQ| ...diagram = leg\;(L) \\[3ex] \implies \\[3ex] HL\;\;Congruency \\[5ex] \underline{Second\;\;Approach:\;\;SSS} \\[3ex] |AC| = |PR| = diameter = hypotenuse = side \;(S) \\[3ex] |AB| = |PQ| ...diagram = side\;(S) \\[3ex] |BC|^2 = |AC|^2 - |AB|^2 ...Pythagorean\;\;Theorem \\[3ex] Similarly \\[3ex] |QR|^2 = |PR|^2 - |PQ|^2 ...Pythagorean\;\;Theorem \\[3ex] \implies \\[3ex] |BC|^2 = |QR|^2 \\[3ex] |BC| = |QR| ...side\;(S) \\[3ex] \implies \\[3ex] SSS\;\;Congruency \\[5ex] \underline{Third\;\;Approach:\;\;SAS} \\[3ex] |AB| = |PQ| ...diagram = side\;(S) \\[3ex] \angle ABC = \angle PQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] |BC|^2 = |AC|^2 - |AB|^2 ...Pythagorean\;\;Theorem \\[3ex] Similarly \\[3ex] |QR|^2 = |PR|^2 - |PQ|^2 ...Pythagorean\;\;Theorem \\[3ex] \implies \\[3ex] |BC|^2 = |QR|^2 \\[3ex] |BC| = |QR| ...side\;(S) \\[3ex] \implies \\[3ex] SAS\;\;Congruency \\[3ex] $(b) William is correct. A kite is a quadrilateral. Any circle can be drawn around a quadrilateral (the four vertices of a quadrilateral). That circle would be a cyclic quadrilateral where the opposite angles are right angles. The sum of those opposite angles are supplementary...opposite angles of a cyclic quadrilateral are supplementary (sum of 180°) (223.) GCSE B, C, D and E are points on a circle. AB is the tangent at B to the circle. AB is parallel to ED Angle ABE = 73° Work out the size of angle DCE Give a reason for each stage of your working. We have a chord and a tangent, so we may need the Alternate Segment Theorem Let us do some construction Construction: (1.) Draw a chord from Point B to Point D ...so we can use the Alternate Segment Theorem (2.) Draw a chord from Point C to Point D ...we may need to use another Theorem$ \angle BED = 73^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BDE = 73^\circ ...\angle\;\;between\;\;tangent\;BA\;\;and\;\;chord\;RL = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DBE + \angle BDE + \angle BED = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BED \\[3ex] \angle DBE + 73 + 73 = 180 \\[3ex] \angle DBE + 146 = 180 \\[3ex] \angle DBE = 34^\circ \\[3ex] \angle DCE = \angle DBE = 34^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $(224.) WASSCE The diagram shows a circle with centre O If ∠STR = 29° and ∠RST = 46°, calculate the value of ∠STO$ A.\;\; 12^\circ \\[3ex] B.\;\; 15^\circ \\[3ex] C.\;\; 29^\circ \\[3ex] D.\;\; 34^\circ \\[3ex]  \angle ROT = 2 * \angle RST \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle ROT = 2 * 46 \\[3ex] \angle ROT = 92^\circ \\[3ex] \angle ORT = \angle OTR = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OTR \\[3ex] \angle ORT + \angle OTR + \angle ROT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ORT \\[3ex] p + p + 92 = 180 \\[3ex] 2p = 180 - 92 \\[3ex] 2p = 88 \\[3ex] p = \dfrac{88}{2} \\[5ex] p = 44^\circ \\[3ex] \therefore \angle OTR = 44^\circ \\[3ex] \angle OTR = \angle STR + \angle STO ...diagram \\[3ex] 44 = 29 + \angle STO \\[3ex] 29 + \angle STO = 44 \\[3ex] \angle STO = 44 - 29 \\[3ex] \angle STO = 15^\circ $(225.) NSC (225.1) Complete the following theorem: Two tangents drawn to a circle from the same point outside the circle ... (225.2) In the diagram below, O is the centre of circle MVS EM and EV are tangents at M and V respectively. MOS is a diameter. Tangent EV and diameter MS are produced to T.$\hat{E} = 52^\circ$(225.2.1) Give a reason why$E\hat{M}O = 90^\circ$(225.2.2) Determine, stating reasons, the size of EACH of the following angles: (a)$\hat{V_1}$(b)$\hat{S_2}$(c)$\hat{O_1}$(d)$\hat{V_4}$(e)$\hat{T}$(225.1) Two tangents drawn to a circle from the same point outside the circle are equal in length.$ (225.2.1) \\[3ex] E\hat{M}O = 90^\circ ...radius\;OM \perp tangent\;EM\;\;at\;\;point\;\;of\;\;contact\;M \\[3ex] (225.2.2) \\[3ex] (a) \\[3ex] |EV| = |EM| ...two\;\;tangents:\;\;EV\;\;and\;\;EM\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;E\;\;are\;\;equal\;\;in\;\;length \\[3ex] \hat{V_1} = \hat{M_1} = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EMV \\[3ex] p + p + 52 = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EMV \\[3ex] 2p = 180 - 52 \\[3ex] 2p = 128 \\[3ex] p = \dfrac{128}{2} \\[5ex] p = 64 \\[3ex] \therefore \hat{V_1} = \hat{M_1} = 64^\circ \\[3ex] (b) \\[3ex] \hat{M_1} + \hat{M_2} = E\hat{M}O = 90^\circ ... diagram \\[3ex] 64 + \hat{M_2} = 90 \\[3ex] \hat{M_2} = 90 - 64 \\[3ex] \hat{M_2} = 26^\circ \\[3ex] \angle SMV = \hat{M_2} = 26^\circ ... diagram \\[3ex] \angle MVS = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle MSV + \angle SMV + \angle MVS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MSV \\[3ex] \angle MSV + 26 + 90 = 180 \\[3ex] \angle MSV + 116 = 180 \\[3ex] \angle MSV = 180 - 116 \\[3ex] \angle MSV = 64 \\[3ex] \hat{S_2} = \angle MSV = 64^\circ ... diagram \\[3ex] (c) \\[3ex] \hat{V_3} = \hat{S_2} = 64^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OVS \\[3ex] \hat{O_1} + \hat{V_3} + \hat{S_2} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OVS \\[3ex] \hat{O_1} + 64 + 64 = 180 \\[3ex] \hat{O_1} + 128 = 180 \\[3ex] \hat{O_1} = 180 - 128 \\[3ex] \hat{O_1} = 52^\circ \\[3ex] (d) \\[3ex] \angle MVT = \hat{E} + \hat{M_1} ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle MVT = 52 + 64 \\[3ex] \angle MVT = 116 \\[3ex] \angle MVT = \angle MVS + \hat{V_4} ... diagram \\[3ex] \implies \\[3ex] \angle MVS + \hat{V_4} = 116 \\[3ex] 90 + \hat{V_4} = 116 \\[3ex] \hat{V_4} = 116 - 90 \\[3ex] \hat{V_4} = 26^\circ \\[3ex] (e) \\[3ex] \hat{S_2} = \hat{V_4} + \hat{T} ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] 64 = 26 + \hat{T} \\[3ex] \hat{T} + 26 = 64 \\[3ex] \hat{T} = 64 - 26 \\[3ex] \hat{T} = 38^\circ $(226.) GCSE (a) The diagram shows a circle with centre O. Write down the word from the box that describes the line AB (b) The diagram shows a circle with centre O On the diagram, draw a radius of the circle. (c) Work out the size of the angle marked x (a) Line AB is a tangent because it touches the circle (circumference of the circle) at only one point. (b)$|OP|$= r = radius$ (c) \\[3ex] x + 130 = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] x = 360 - 130 \\[3ex] x = 230^\circ $(227.) GCSE Points B, D, E and F lie on a circle. ABC is the tangent to the circle at B. Find the size of angle ABD. You must give a reason for each stage of your working.$ \angle CBF = \angle BDF = 40^\circ ...\angle\;\;between\;\;tangent\;ABC\;\;and\;\;chord\;BF = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DBF + \angle DEF = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle DBF + 100 = 180 \\[3ex] \angle DBF = 180 - 100 \\[3ex] \angle DBF = 80^\circ \\[3ex] \angle ABD + \angle DBF + \angle CBF = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ABD + 80 + 40 = 180 \\[3ex] \angle ABD + 120 = 180 \\[3ex] \angle ABD = 180 - 120 \\[3ex] \angle ABD = 60^\circ $(228.) WASSCE The diagram shows a circle with centre O If ∠ZYW = 33°, find ∠ZWX$ A.\;\; 33^\circ \\[3ex] B.\;\; 57^\circ \\[3ex] C.\;\; 90^\circ \\[3ex] D.\;\; 100^\circ \\[3ex]  \angle ZXW = \angle ZYW = 33^\circ \\[3ex] ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle WZX = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ZWX + \angle WZX + \angle ZXW = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WZX \\[3ex] \angle ZWX + 90 + 33 = 180 \\[3ex] \angle ZWX + 123 = 180 \\[3ex] \angle ZWX = 180 - 123 \\[3ex] \angle ZWX = 57^\circ $(229.) NSC In the diagram, PQRS is a cyclic quadrilateral. PS is produced to W. TR and TS are tangents to the circle at R and S respectively.$\hat{T} = 78^\circ$and$\hat{Q} = 93^\circ$(229.1) Give a reason why ST = TR (229.2) Calculate, giving reasons, the size of: (229.2.1)$\hat{S_2}$(229.2.2)$\hat{S_3} (229.1) \\[3ex] ST = TR ...two\;\;tangents:\;\;ST\;\;and\;\;TR\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] (229.2) \\[3ex] (229.2.1) \\[3ex] \hat{S_2} = \hat{R_2} = k ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle TRS \\[3ex] \hat{S_2} + \hat{R_2} + \hat{T} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TRS \\[3ex] k + k + 78 = 180 \\[3ex] 2k = 180 - 78 \\[3ex] 2k = 102 \\[3ex] k = \dfrac{102}{2} \\[5ex] k = 51 \\[3ex] \therefore \hat{S_2} = \hat{R_2} = 51^\circ \\[3ex] (229.2.2) \\[3ex] \hat{S_2} + \hat{S_3} = \hat{Q} ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] 51 + \hat{S_3} = 93 \\[3ex] \hat{S_3} = 93 - 51 \\[3ex] \hat{S_3} = 42^\circ $(230.) GCSE The diagram below shows a circle with centre at point O. A, B, C and D are all points on the circumference of the circle. AB = 7.5cm and BC = 4.7cm. (a) (i) Give the reason why$A\hat{B}C$is 90° (ii) Calculate the size of angle x (b) Write down the size of angle y State the circle theorem you have used to find your answer.$ (a)\;\;(i) \\[3ex] A\hat{B}C = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (ii) \\[3ex] \underline{\triangle ABC} \\[3ex] SOHCAHTOA \\[3ex] \tan x = \dfrac{opp}{adj} \\[5ex] \tan x = \dfrac{7.5}{4.7} \\[5ex] \tan x = 1.595744681 \\[3ex] x = \tan^{-1} (1.595744681) \\[3ex] x = 57.92599912 \\[3ex] x \approx 57.9^\circ \\[3ex] (b) \\[3ex] y = x = 57.92599912 ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] y \approx 57.9^\circ $(231.) GCSE A, B, C and D are points on a circle. PCQ is a tangent to the circle. AB = CB Angle BCQ = x° Prove that angle CDA = 2x° Give reasons for each stage in your working. Construction: Draw a chord from Point C to Point A$ \angle BAC = x^\circ ...\angle\;\;between\;\;tangent\;PCQ\;\;and\;\;chord\;RL = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle ACB = \angle BAC = x^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle CBA \\[3ex] \angle CBA + \angle ACB + \angle BAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CBA \\[3ex] \angle CBA + x + x = 180 \\[3ex] \angle CBA + 2x = 180 \\[3ex] \angle CBA = 180 - 2x \\[5ex] \angle CDA + \angle CBA = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle CDA = 180 - \angle CBA \\[3ex] \angle CDA = 180 - (180 - 2x) \\[3ex] \angle CDA = 180 - 180 + 2x \\[3ex] \angle CDA = 2x^\circ $(232.) WASSCE In the diagram, PQ and PS are tangents to the circle centre O If ∠PSQ = m, ∠SPQ = n and ∠SQR = 33°, find the value of (m + n)$ A.\;\; 103^\circ \\[3ex] B.\;\; 123^\circ \\[3ex] C.\;\; 133^\circ \\[3ex] D.\;\; 143^\circ \\[3ex] $We can solve this question using at least two approaches Use whatever approach you prefer.$ \underline{First\;\;Approach} \\[3ex] $Join PQ to K$ \angle OQK = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle SQK = 33 + 90 ...diagram \\[3ex] \angle SQK = 123^\circ \\[3ex] m + n = \angle SQK = 123^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle OQP = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle SQP + \angle SQR = \angle OQP ...diagram \\[3ex] \angle SQP + 33 = 90 \\[3ex] \angle SQP = 90 - 33 \\[3ex] \angle SQP = 57^\circ \\[3ex] m + n + \angle SQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] m + n + 57 = 180 \\[3ex] m + n = 180 - 57 \\[3ex] m + n = 123^\circ $(233.) WASSCE In the figure, PQ is a tangent to the circle LMR at R If LM is a diameter and ∠RLM = 35°, calculate the value of x$ A.\;\; 75^\circ \\[3ex] B.\;\; 65^\circ \\[3ex] C.\;\; 55^\circ \\[3ex] D.\;\; 45^\circ \\[3ex]  \angle MRL = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] x = \angle RML ...\angle\;\;between\;\;tangent\;PRQ\;\;and\;\;chord\;RL = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle RML + \angle RLM + \angle MRL = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MRL \\[3ex] x + 35 + 90 = 180 \\[3ex] x + 125 = 180 \\[3ex] x = 180 - 125 \\[3ex] x = 55^\circ $(234.) GCSE B, C, D and F are points on a circle. ABC, AFD, BFE and CDE are straight lines. Work out the size of angle x Show your working clearly.$ \underline{Cyclic\;\;Quadrilateral\;BFDC} \\[3ex] x + \angle ADC = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle ADC = 180 - x \\[5ex] \underline{\triangle BEC} \\[3ex] \angle BCE + x + 32 = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BEC \\[3ex] \angle BCE = 180 - x - 32 \\[3ex] \angle BCE = 148 - x \\[3ex] \angle ACD = \angle BCE = 148 - x ...diagram \\[5ex] \underline{\triangle ACD} \\[3ex] 54 + \angle ADC + \angle ACD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] 54 + (180 - x) + (148 - x) = 180 \\[3ex] 54 + 180 - x + 148 - x = 180 \\[3ex] 382 - 2x = 180 \\[3ex] 382 - 180 = 2x \\[3ex] 202 = 2x \\[3ex] 2x = 202 \\[3ex] x = \dfrac{202}{2} \\[5ex] x = 101^\circ $(235.) NSC In the diagram, chords DE, EF and DF are drawn in the circle with centre O. KFC is a tangent to the circle at F. Prove the theorem which states that$D\hat{F}K = \hat{E}$Construction: Join the radii: from centre O to the point D, and from centre O to point F Radii: |OD| and |OF| This forms a triangle. Label the angles in the triangle.$ \angle ODF = \theta \\[3ex] \angle OFD = \theta \\[3ex] \angle DOF = \psi \\[5ex] \angle ODF = \angle OFD = \theta ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DOF \\[3ex] \theta + \theta + \psi = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle DOF \\[3ex] 2\theta + \psi = 180 \\[3ex] \psi = 180 - 2\theta ...eqn.(1) \\[5ex] \psi = 2 * \hat{E} ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] \psi = 2\hat{E} ...eqn.(2) \\[5ex] \psi = \psi \implies eqn.(2) = eqn.(1) \\[3ex] 2\hat{E} = 180 - 2\theta \\[3ex] 2\hat{E} = 2(90 - \theta) \\[3ex] \hat{E} = 90 - \theta \\[3ex] \theta = 90 - \hat{E} ... eqn.(3) \\[5ex] \theta + D\hat{F}K = 90^\circ ...radius\;OF \perp tangent\;KFC\;\;at\;\;point\;\;contact\;F \\[3ex] \theta + D\hat{F}K = 90 \\[3ex] \theta = 90 - D\hat{F}K ...eqn.(4) \\[5ex] \theta = \theta \implies eqn.(3) = eqn.(4) \\[3ex] 90 - \hat{E} = 90 - D\hat{F}K \\[3ex] D\hat{F}K = 90 - 90 + \hat{E} \\[3ex] D\hat{F}K = \hat{E} \\[3ex] $Therefore, the angle between the tangent KFC and the chord DF (angle DFK) is equal to the angle in the alternate segment (angle E) (236.) WASSCE In the diagram, ∠RQS = 40°, |RT| = |PT| and ∠RMS = y Find the value of y$ \angle MTS = 40^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle PTR + \angle MTS = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle PTR + 40 = 180 \\[3ex] \angle PTR = 180 - 40 \\[3ex] \angle PTR = 140^\circ \\[3ex] \angle TPR = \angle TRP ...base\;\; \angle s\;\;of\;\;isosceles\;\;\triangle TPR \\[3ex] y = \angle TRP + \angle TRP \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle TPR + \angle TRP + \angle PTR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TPR \\[3ex] \implies \\[3ex] y + 140 = 180 \\[3ex] y = 180 - 140 \\[3ex] y = 40^\circ $(237.) WASSCE XY is a tangent to a circle LMN at the point M XLN is a straight line. ∠NXM = 34° and ∠NMY = 65° (i.) Illustrate the information in a diagram (ii.) Find the value of: (α.) ∠MLX (β.) ∠LNM (i) The diagram for the information is:$ (ii) \\[3ex] \angle MLN = 65^\circ ...\angle\;\;between\;\;tangent\;XMY\;\;and\;\;chord\;MN = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (\alpha.) \\[3ex] \angle MLX + \angle MLN = 180^\circ \\[3ex] \angle MLX + 65 = 180 \\[3ex] \angle MLX = 180 - 65 \\[3ex] \angle MLX = 115^\circ \\[3ex] (\beta.) \\[3ex] \angle NMX + \angle NMY = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle NMX + 65 = 180 \\[3ex] \angle NMX = 180 - 65 \\[3ex] \angle NMX = 115^\circ \\[3ex] \angle XNM + \angle NXM + \angle NMX = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XNM \\[3ex] \angle XNM + 34 + 115 = 180 \\[3ex] \angle XNM + 149 = 180 \\[3ex] \angle XNM = 180 - 149 \\[3ex] \angle XNM = 31^\circ \\[3ex] \angle LNM = \angle XNM = 31^\circ...diagram $(238.) GCSE M, N and P are points on a circle, centre O MON is a diameter of the circle. MP = 3.5cm PN = 9.7cm Angle MPN = 90° Work out the circumference of the circle. Give your answer correct to 3 significant figures.$ |MN|^2 = |MP|^2 + |PN|^2 ...Pythagoras\;\;Theorem \\[3ex] |MN|^2 = 3.5^2 + 9.7^2 \\[3ex] |MN|^2 = 12.25 + 94.09 \\[3ex] |MN|^2 = 106.34 \\[3ex] |MN| = \sqrt{106.34} \\[3ex] |MN| = 10.31212878 \\[3ex] diameter = d = |MN| = 10.31212878\;cm \\[3ex] circumference = C \\[3ex] C = \pi d \\[3ex] C = \dfrac{22}{7} * 10.31212878 \\[5ex] C = \dfrac{226.8668332}{7} \\[5ex] C = 32.4095476 \\[3ex] C \approx 32.4\;cm $(239.) GCSE A and B are points on the circumference of a circle, centre O. AC is a tangent to the circle. Angle BAC = 2x Find the size of the angle AOB, in terms of x, giving a reason for each stage of your working. We can solve this question using at least two approaches Use any approach you prefer. First Approach$ \angle OAC = 90^\circ ...radius\;OA \perp tangent\;AC\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAC = \angle OAB + \angle BAC ...diagram \\[3ex] 90 = \angle OAB + 2x \\[3ex] \angle OAB + 2x = 90 \\[3ex] \angle OAB = 90 - 2x \\[3ex] \angle OBA = \angle OAB = 90 - 2x ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle \\[3ex] \angle AOB + \angle OAB + \angle OBA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB + (90 - 2x) + (90 - 2x) = 180 \\[3ex] \angle AOB + 90 - 2x + 90 - 2x = 180 \\[3ex] \angle AOB + 180 - 4x = 180 \\[3ex] \angle AOB = 180 - 180 + 4x \\[3ex] \angle AOB = 4x^\circ \\[3ex] $Second Approach Let us do a little work on the diagram so we can solve the question. Construction: Draw a chord from point A and another chord from point B, to meet at a point say E on the circumference$ \angle AEB = \angle BAC = 2x ...\angle\;\;between\;\;tangent\;AC\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle AOB = 2 * \angle AEB ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOB = 2 * 2x \\[3ex] \angle AOB = 4x^\circ $(240.) ICSE In the figure given below, O is the centre of the circle and AB is a diameter. If AC = BD = ∠AOC = 72°, find: (i) ∠ABC (ii) ∠BAD (iii) ∠ABD$ (i) \\[3ex] \angle OAC = \angle OCA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOC \\[3ex] \angle OAC + \angle OCA + \angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] p + p + 72 = 180 \\[3ex] 2p = 180 - 72 \\[3ex] 2p = 108 \\[3ex] p = \dfrac{108}{2} \\[5ex] p = 54 \\[3ex] \therefore \angle OAC = \angle OCA = 54^\circ \\[3ex] \angle BAC = \angle OAC = 54^\circ ... diagram \\[3ex] \angle BCA = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ABC + \angle BAC + \angle BCA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle ABC + 54 + 90 = 180 \\[3ex] \angle ABC + 144 = 180 \\[3ex] \angle ABC = 180 - 144 \\[3ex] \angle ABC = 36^\circ \\[3ex] (ii) \\[3ex] \angle ACB = \angle BDA = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \\[3ex] \triangle ACB \;\;and\;\; \triangle BDA \;\;are\;\;right\;\;\triangle s \\[3ex] |AC| = |BD| ...given...leg\;(L) \\[3ex] |AB| = |AB| ...diameter = hypotenuse\;(H) \\[3ex] \therefore \triangle ACB \cong \triangle BDA ... HL\;\;Congruency \\[3ex] \implies \\[3ex] \angle BAD = \angle ABC = 36^\circ ... \triangle ACB \cong \triangle BDA \\[3ex] (iii) \\[3ex] \angle ABD + \angle BAD + \angle BDA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \angle ABD + 36 + 90 = 180 \\[3ex] \angle ABD + 126 = 180 \\[3ex] \angle ABD = 180 - 126 \\[3ex] \angle ABD = 54^\circ $(241.) ICSE In the given figure TP and TQ are two tangents to the circle with centre O, touching at A and C respectively. If ∠BCQ = 55° and ∠BAP = 60°, find: (i) ∠OBA and ∠OBC (ii) ∠AOC (iii) ∠ATC Construction: Draw a chord from point A to point C$ (i) \\[3ex] \angle OAT = 90^\circ ...radius\;OA \perp tangent\;PAT\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle BAP + \angle OAB + \angle OAT = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 60 + \angle OAB + 90 = 180 \\[3ex] \angle OAB + 150 = 180 \\[3ex] \angle OAB = 180 - 150 \\[3ex] \angle OAB = 30^\circ \\[3ex] \angle OBA = \angle OAB = 30^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OBA \\[5ex] Also: \\[3ex] \angle OCT = 90^\circ ...radius\;OC \perp tangent\;QCT\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] \angle OCT + \angle OCB + \angle BCQ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + \angle OCB + 55 = 180 \\[3ex] \angle OCB + 145 = 180 \\[3ex] \angle OCB = 180 - 145 \\[3ex] \angle OCB = 35^\circ \\[3ex] \angle OBC = \angle OCB = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OBC \\[3ex] (ii) \\[3ex] \angle ABC = \angle OBA + \angle OBC ... diagram \\[3ex] \angle ABC = 30 + 35 \\[3ex] \angle ABC = 65^\circ \\[3ex] \angle AOC = 2 * \angle ABC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOC = 2 * 65 \\[3ex] \angle AOC = 130^\circ \\[3ex] (iii) \\[3ex] We\;\;can\;\;solve\;\;this\;\;question\;\;using\;\;at\;\;least\;\;two\;\;approaches \\[3ex] Use\;\;any\;\;approach\;\;you\;\;prefer \\[3ex] \underline{First\;\;Approach} \\[3ex] \angle OAC = \angle OCA = m ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOC \\[3ex] \angle OAC + \angle OCA + \angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] m + m + 130 = 180 \\[3ex] 2m = 180 - 130 \\[3ex] 2m = 50 \\[3ex] m = \dfrac{50}{2} \\[5ex] m = 25 \\[3ex] \therefore \angle OAC = \angle OCA = 25^\circ \\[3ex] \angle OAT = \angle OAC + \angle CAT ... diagram \\[3ex] 90 = 25 + \angle CAT \\[3ex] 25 + \angle CAT = 90 \\[3ex] \angle CAT = 90 - 25 \\[3ex] \angle CAT = 65^\circ \\[3ex] |AT| = |CT| ...two\;\;tangents:\;\;AT\;\;and\;\;CT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \angle ACT = \angle CAT = 65^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ATC \\[3ex] \angle ATC + \angle ACT + \angle CAT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ATC \\[3ex] \angle ATC + 65 + 65 = 180 \\[3ex] \angle ATC + 130 = 180 \\[3ex] \angle ATC = 180 - 130 \\[3ex] \angle ATC = 50^\circ \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle ATC + \angle BAT + \angle ABC + \angle BCT = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;BCTA \\[3ex] \angle BAT = \angle OAB + \angle OAT ... diagram \\[3ex] \angle BAT = 30 + 90 \\[3ex] \angle BAT = 120^\circ \\[3ex] \angle BCT = \angle OCB + \angle OCQ ... diagram \\[3ex] \angle BAT = 35 + 90 \\[3ex] \angle BAT = 125^\circ \\[3ex] \implies \\[3ex] \angle ATC + 120 + 65 + 125 = 360 \\[3ex] \angle ATC + 310 = 360 \\[3ex] \angle ATC = 360 - 310 \\[3ex] \angle ATC = 50^\circ $(242.) GCSE Figure 1 shows the circle ABCD with centre O and diameter DC The point T is such that TCOD is a straight line and TA is the tangent to the circle at A AT = 10cm TC = 8cm (a) Calculate the radius, in cm, of the circle. (b) Calculate the length, in cm to 3 significant figures, of the arc ABC$ (a) \\[3ex] \underline{Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] \overline{TD} * 8 = 10 * 10 \\[3ex] \overline{TD} = \dfrac{10 * 10}{8} \\[5ex] \overline{TD} = 12.5 \\[3ex] But: \\[3ex] \overline{TD} = \overline{CO} + \overline{OD} + \overline{TC}...diagram \\[3ex] \overline{CO} = \overline{OD} = radius = r ...diagram \\[3ex] \implies \\[3ex] \overline{DT} = r + r + 8 \\[3ex] 12.5 = 2r + 8 \\[3ex] 2r + 8 = 12.5 \\[3ex] 2r = 12.5 - 8 \\[3ex] 2r = 4.5 \\[3ex] r = \dfrac{4.5}{2} \\[5ex] r = 2.25cm \\[3ex] $(b) Let us visualize what we need to find First, we have to find the angle, &theta Then, we use the formula for the length of an arc to find it$ \angle OAT = 90^\circ \\[3ex] ...radius\;OA \perp tangent\;AT\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] SOHCAHTOA \\[3ex] \sin \theta = \dfrac{opp}{hyp} = \dfrac{|AT|}{|TO|} \\[5ex] |TO| = |TC| + |CO| \\[3ex] |TO| = 8 + 2.25 = 10.25cm \\[3ex] \implies \\[3ex] \sin \theta = \dfrac{10}{10.25} \\[5ex] \sin \theta = 0.9756097561 \\[3ex] \theta = \sin^{-1}(0.9756097561) \\[3ex] \theta = 77.31961651^\circ \\[3ex] length\;\;of\;\;arc = L \\[3ex] L = \dfrac{\theta}{360} * 2\pi r \\[5ex] L = \dfrac{77.31961651}{360} * 2 * \dfrac{22}{7} * 2.25 \\[5ex] L = \dfrac{7654.642034}{2520} \\[5ex] L = 3.037556363 \\[3ex] L \approx 3.04cm...to\;\;3\;\;significant\;\;figures $(243.) GCSE The diagram shows a circle, centre O. Points A, B and C lie on the circumference of the circle. Line AOB is a diameter. Line DAE is a tangent to the circle. Angle CAE = 32° (a) Give a reason why angle ACB is a right angle. (b) The radius of the circle is 8cm. Calculate length BC.$ (a) \\[3ex] \angle ACB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (b) \\[3ex] \angle BAC + 32 = 90^\circ ...radius\;OA \perp tangent\;DAE\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle BAC = 90 - 32 \\[3ex] \angle BAC = 58^\circ \\[3ex] radius = |OA| = |OB| = 8\;cm \\[3ex] diameter = |AB| = 2(8) = 16\;cm \\[3ex] \dfrac{|BC|}{\sin \angle BAC} = \dfrac{|AB|}{\sin \angle ACB} ... Sine\;\;Rule \\[5ex] \dfrac{|BC|}{\sin 58} = \dfrac{16}{\sin 90} \\[5ex] |BC| = \dfrac{16 \sin 58}{\sin 90} \\[5ex] |BC| = \dfrac{16(0.8480480962)}{1} \\[5ex] |BC| = 13.56876954 \\[3ex] |BC| \approx 13.6\;cm $(244.) WASSCE In the diagram, O is the centre of the circle, POR is a diameter and ∠PQS = 37° What is ∠SPR?$ A.\;\; 147^\circ \\[3ex] B.\;\; 127^\circ \\[3ex] C.\;\; 55^\circ \\[3ex] D.\;\; 37^\circ \\[3ex]  \angle PRS = 37^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle RSP = 90^\circ...\angle\;in\;\;a\;\;semicircle \\[3ex] \angle SPR + \angle PRS + \angle RSP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] \angle SPR + 37 + 90 = 180 \\[3ex] \angle SPR + 127 = 180 \\[3ex] \angle SPR = 180 - 127 \\[3ex] \angle SPR = 53^\circ $(245.) WASSCE In the diagram, RQ and PQ are tangents to the circle centre O and ∠PQR = 80° Calculate the reflex ∠POR$ A.\;\; 160^\circ \\[3ex] B.\;\; 200^\circ \\[3ex] C.\;\; 260^\circ \\[3ex] D.\;\; 310^\circ \\[3ex]  \angle ORQ = 90^\circ ...radius\;OR \perp tangent\;RQ\;\;at\;\;point\;\;of\;\;contact\;R \\[3ex] \angle OPQ = 90^\circ ...radius\;OP \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;P \\[3ex] \angle OQR = \dfrac{80}{2} = 40^\circ ... two\;\;tangents\;\;drawn\;\;from\;\;same\;\;external\;\;point\;\;bisects\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents \\[5ex] \angle OQP = \dfrac{80}{2} = 40^\circ ... two\;\;tangents\;\;drawn\;\;from\;\;same\;\;external\;\;point\;\;bisects\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents \\[5ex] \angle ROQ + \angle ORQ + \angle OQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROQ \\[3ex] \angle ROQ + 90 + 40 = 180 \\[3ex] \angle ROQ + 130 = 180 \\[3ex] \angle ROQ = 180 - 130 \\[3ex] \angle ROQ = 50^\circ \\[3ex] \angle POQ = \angle ROQ = 50^\circ ... two\;\;tangents\;\;drawn\;\;from\;\;same\;\;external\;\;point\;\;bisects\;\;\angle\;\;formed\;\;by\;\;the\;\;radii \\[5ex] Obtuse\;\angle POR = \angle POQ + \angle ROQ ... diagram \\[3ex] = 50 + 50 \\[3ex] = 100^\circ \\[3ex] Reflex\;\angle POR + Obtuse\;\angle POR = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\angle POR + 100 = 360 \\[3ex] = 360 - 100 \\[3ex] = 260^\circ $(246.) NSC In the diagram below, ABCD is a cyclic quadrilateral. AB and DC are produced to meet at E. AD and BC are produced to meet at F.$A\hat{F}B = 2x$,$D\hat{A}B = 3x$and$A\hat{E}D = x$Determine, giving reasons, the value of$x \underline{\triangle BAF} \\[3ex] \hat{B_2} + 3x + 2x = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \hat{B_2} + 5x = 180 \\[3ex] \hat{B_2} = 180 - 5x \\[5ex] \underline{\triangle EAD} \\[3ex] \hat{D_2} + 3x + x = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \hat{D_2} + 4x = 180 \\[3ex] \hat{D_2} = 180 - 4x \\[5ex] \hat{B_2} + \hat{D_2} = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \implies \\[3ex] (180 - 5x) + (180 - 4x) = 180 \\[3ex] 180 - 5x + 180 - 4x = 180 \\[3ex] 360 - 9x = 180 \\[3ex] 360 - 180 = 9x \\[3ex] 180 = 9x \\[3ex] 9x = 180 \\[3ex] x = \dfrac{180}{9} \\[5ex] x = 20^\circ $(247.) GCSE A, B, C and D are points on the circumference of the circle. The line XY is a tangent to the circle at A. (a) Find the value of x, giving a reason for your answer. (b) Find the value of y, giving a reason for your answer.$ (a) \\[3ex] x = 55^\circ ... \angle\;\;in\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;XAY\;\;and\;\;chord\;AD \\[3ex] (b) \\[3ex] y + 65 = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] y = 180 - 65 \\[3ex] y = 115^\circ $(248.) WASSCE If ∠VWZ = 110°, find ∠WXZ$ A.\;\; 110^\circ \\[3ex] B.\;\; 100^\circ \\[3ex] C.\;\; 85^\circ \\[3ex] D.\;\; 55^\circ \\[3ex]  \angle WXZ = \angle WZX = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle WXZ \\[3ex] \angle VWZ = \angle WXZ + \angle WZX \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;\angle s \\[3ex] 110 = p + p \\[3ex] 2p = 110 \\[3ex] p = \dfrac{110}{2} \\[5ex] p = 55 \\[3ex] \therefore \angle WXZ = 55^\circ $(249.) WASSCE In the diagram, PS || QR, ∠PSR = 79°, ∠SPR = 54° and TQ is a tangent to the circle at P Calculate: (i) ∠TPS (ii) ∠PQR$ \angle SRP + \angle SPR + \angle PSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] \angle SRP + 54 + 79 = 180 \\[3ex] \angle SRP + 133 = 180 \\[3ex] \angle SRP = 180 - 133 \\[3ex] \angle SRP = 47^\circ \\[3ex] (i) \\[3ex] \angle TPS = \angle SRP = 47^\circ ... \angle\;\;between\;\;tangent\;TPQ\;\;and\;\;chord\;PS = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (ii) \\[3ex] \angle TPS + \angle SPR + \angle RPQ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 47 + 54 + \angle RPQ = 180 \\[3ex] 101 + \angle RPQ = 180 \\[3ex] \angle RPQ = 180 - 101 \\[3ex] \angle RPQ = 79^\circ \\[3ex] \angle PRQ = 54^\circ ... alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle PQR + \angle PRQ + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRQ \\[3ex] \angle PQR + 54 + 79 = 180 \\[3ex] \angle PQR + 133 = 180 \\[3ex] \angle PQR = 180 - 133 \\[3ex] \angle PQR = 47^\circ $(250.) GCSE A, C, B and D are four points on a circle. The chord AB intersects the chord CD at P AP = 7cm PB = 5cm PD = 4cm Calculate, in cm, the length of CP$ CP * PD = AP * PB ... Intersecting\;\;Chords\;\;Theorem \\[3ex] CP * 4 = 7 * 5 \\[3ex] CP = \dfrac{7 * 5}{4} \\[5ex] CP = 8.75 \\[3ex] |CP| = 8.75cm $(251.) ICSE In the given figure AB = 9 cm, PA = 7.5 cm and PC = 5 cm. Chords AD and BC intersect at P. (i) Prove that$\triangle PAB \sim \triangle PCD$(ii) Find the length of CD. (iii) Find area of$\triangle PAB$:$\triangle PCD (i) \\[3ex] \angle B = \angle D ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal = Angle\;(A) \\[3ex] \angle A = \angle C ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal = Angle\;(A) \\[3ex] \therefore \triangle PAB \sim \triangle PCD ...AA\;\;Similarity\;\;Postulate \\[3ex] (ii) \\[3ex] \dfrac{|CD|}{|AB|} = \dfrac{|PC|}{|PA|} \\[5ex] \dfrac{|CD|}{9} = \dfrac{5}{7.5} \\[5ex] |CD| = \dfrac{9(5)}{7.5} \\[5ex] |CD| = 6\;cm \\[3ex] (iii) \\[3ex] Area\;\;of\;\;\triangle PAB \\[3ex] = \dfrac{1}{2} * |PA| * |AB| * \sin \angle A \\[5ex] = 0.5(7.5)(9)\sin \angle A \\[3ex] Area\;\;of\;\;\triangle PCD \\[3ex] = \dfrac{1}{2} * |PC| * |DC| * \sin \angle C \\[5ex] = 0.5(5)(6)\sin \angle C \\[3ex] Ratio\;\;of\;\;Areas \\[3ex] = \dfrac{Area\;\;of\;\;\triangle PAB}{Area\;\;of\;\;\triangle PCD} \\[5ex] = \dfrac{0.5(7.5)(9)\sin \angle A}{0.5(5)(6)\sin \angle C} \\[5ex] \angle A = \angle C \implies \sin \angle A = \sin \angle C \\[3ex] \implies \\[3ex] = \dfrac{22.5}{10} \\[5ex] = \dfrac{225}{100} \\[5ex] = \dfrac{9}{4} \\[5ex] = 9 : 4 $(252.) WASSCE In the diagram, PQRS is a cyclic quadrilateral. If |SR| = |RQ|, ∠SRP = 60° and ∠RPQ = 48°, find ∠PRQ$ \angle QSR = 48^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle SQR = \angle QSR = 48^\circ ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RSQ \\[3ex] \angle SRQ = \angle SRP + \angle PRQ ... diagram \\[3ex] \angle SRQ + \angle QSR + \angle SQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RSQ \\[3ex] \implies \\[3ex] (\angle SRP + \angle PRQ) + \angle QSR + \angle SQR = 180 \\[3ex] 60 + \angle PRQ + 48 + 48 = 180 \\[3ex] \angle PRQ + 156 = 180 \\[3ex] \angle PRQ = 180 - 156 \\[3ex] \angle PRQ = 24^\circ $(253.) WASSCE In the diagram, LMT is a straight line. If O is the centre of the circle LMN, ∠OMN = 20°, ∠LTN = 32° and |NM| = |MT|, find ∠LNM$ A.\;\; 44^\circ \\[3ex] B.\;\; 46^\circ \\[3ex] C.\;\; 52^\circ \\[3ex] D.\;\; 76^\circ \\[3ex]  \angle ONM = \angle OMN = 20^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OMN \\[3ex] \angle NOM + \angle ONM + \angle OMN = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle NOM \\[3ex] \angle NOM + 20 + 20 = 180 \\[3ex] \angle NOM + 40 = 180 \\[3ex] \angle NOM = 180 - 40 \\[3ex] \angle NOM = 140^\circ \\[3ex] \angle NOM = 2 * \angle NLM ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 140 = 2 * \angle NLM \\[3ex] 2 * \angle NLM = 140 \\[3ex] \angle NLM = \dfrac{140}{2} \\[5ex] \angle NLM = 70^\circ \\[5ex] \angle MTN = \angle LTN = 32^\circ ...diagram \\[3ex] \angle MNT = \angle MTN = 32^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MNT \\[3ex] \angle LMN = \angle MNT + \angle MTN ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle LMN = 32 + 32 = 64^\circ \\[5ex] \angle LNM + \angle NLM + \angle LMN = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle NLM \\[3ex] \angle LNM + 70 + 64 = 180 \\[3ex] \angle LNM + 134 = 180 \\[3ex] \angle LNM = 180 - 134 \\[3ex] \angle LNM = 46^\circ $(254.) GCSE A, B, C and D are points on a circle, centre O Angle ADC = 132° Calculate, in degrees, the size of angle x$ Reflex\;\;\angle AOC = 2 * \angle ADC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle AOC = 2 * 132 \\[3ex] Reflex\;\;\angle AOC = 264^\circ \\[3ex] x + Reflex\;\;\angle AOC = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] x + 264 = 360 \\[3ex] x = 360 - 264 \\[3ex] x = 96^\circ $(255.) ICSE In the given figure AC is a tangent to the circle with centre O If ∠ADB = 55°, find x and y Give reasons for your answers.$ \angle OAC = 90^\circ ...radius\;OA \perp tangent\;AC\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAC = \angle BAD = \angle OAD = \angle BAC = 90^\circ ... diagram \\[3ex] \angle AOC = 2 * \angle ABD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] y = 2 * \angle ABD \\[3ex] 2 * \angle ABD = y \\[3ex] \angle ABD = \dfrac{y}{2} \\[5ex] \underline{\triangle ABD} \\[3ex] \angle ABD + \angle BAD + \angle ADB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \dfrac{y}{2} + 90 + 55 = 180 \\[5ex] \dfrac{y}{2} + 145 = 180 \\[5ex] \dfrac{y}{2} = 180 - 145 \\[5ex] \dfrac{y}{2} = 35 \\[5ex] y = 2(35) \\[3ex] y = 70^\circ \\[3ex] \underline{\triangle AOC} \\[3ex] \angle ACO + \angle AOC + \angle OAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAC \\[3ex] x + y + 90 = 180 \\[3ex] x + 70 + 90 = 180 \\[3ex] x + 160 = 180 \\[3ex] x = 180 - 160 \\[3ex] x = 20^\circ $(256.) WASSCE In the diagram; O is the centre of the circle |TS| = |SR|, ∠TPR = x°, ∠TQR = y°, ∠TOR = z° and ∠TSR = 118° (i.) Find the relationship between x, y and z (ii.) Calculate ∠STP$ (i) \\[3ex] x = y ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle TOP + z = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle TOP = 180 - z \\[3ex] \angle PTO = \angle TPO = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle TOP \\[3ex] \underline{\triangle TOP} \\[3ex] \angle TPO + \angle PTO + \angle TOP = 180^\circ ...sum\;\;of\;\;\angle s \;\;in\;\;a\;\;\triangle \\[3ex] x + x + (180 - z) = 180 \\[3ex] But\;\; x = y \\[3ex] And\;\;we\;\;need\;\;a\;\;relationship\;\;between\;\;x,\;y,\;z \\[3ex] x + y + 180 - z = 180 \\[3ex] x + y - z = 180 - 180 \\[3ex] x + y - z = 0 \\[3ex] x + y = 0 + z \\[3ex] x + y = z \\[3ex] (ii) \\[3ex] $Construction: Draw the chord from point T to point R$ \angle STR = \angle SRT = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle STR \\[3ex] \angle STR + \angle SRT + \angle TSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle STR \\[3ex] p + p + 118 = 180 \\[3ex] 2p = 180 - 118 \\[3ex] 2p = 62 \\[3ex] p = \dfrac{62}{2} \\[5ex] p = 31 \\[3ex] \therefore \angle STR = 31^\circ \\[3ex] \angle PTR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle STP = \angle STR + \angle PTR ...diagram \\[3ex] \angle STP = 31 + 90 \\[3ex] \angle STP = 121^\circ $(257.) ICSE In the given figure PQ is a tangent to the circle at A, AB and AD are bisectors of ∠CAQ and ∠PAC If ∠BAQ = 30°, prove that: (i) BD is a diameter of the circle (ii) ABC is an isosceles triangle$ \angle BAQ = 30^\circ...given \\[3ex] \implies \\[3ex] \angle CAB = 30^\circ ... AB\;\;bisects\;\;\angle CAQ \\[3ex] \angle CDB = \angle CAB = 30^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle BAQ = \angle ADB = 30^\circ ...\angle\;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle ACB = \angle ADB = 30^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ABC + \angle ACB + \angle CAB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle ABC + 30 + 30 = 180 \\[3ex] \angle ABC + 60 = 180 \\[3ex] \angle ABC = 180 - 60 \\[3ex] \angle ABC = 120^\circ \\[3ex] \angle PAC = \angle ABC = 120^\circ ...\angle\;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AC = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle PAC = \angle PAD + \angle DAC ... diagram \\[3ex] \angle PAD = \angle DAC = \dfrac{1}{2} * 120 = 60^\circ ... AD\;\;bisects\;\;\angle PAC \\[5ex] (i) \\[3ex] \underline{\triangle ABD} \\[3ex] \angle DAB = \angle DAC + \angle CAB ...diagram \\[3ex] \angle DAB = 60 + 30 \\[3ex] \angle DAB = 90^\circ \\[3ex] \implies \\[3ex] |BD| \;\;is\;\;a\;\;diameter \;\;because \angle DAB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (ii) \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ACB = \angle CAB = 30^\circ \\[3ex] \angle ABC = 60^\circ \\[3ex] \implies \\[3ex] \triangle ABC \;\;is\;\;an\;\;isosceles\;\;\triangle $(258.) NYSED Regents Examination In the diagram below of circle O, secant$\overline{ABC}$and tangent$\overline{AD}$are drawn. If CA = 12.5 and CB = 4.5, determine and state the length of$\overline{DA} \overline{BA} = \overline{CA} - \overline{CB} ...diagram \\[3ex] \overline{BA} = 12.5 - 4.5 \\[3ex] \overline{BA} = 8 \\[3ex] \overline{DA}^2 = \overline{BA} * \overline{CA} ...Intersecting\;\;Tangent-Secant\;\;Theorem \\[3ex] \overline{DA}^2 = 8 * 12.5 \\[3ex] \overline{DA}^2 = 100 \\[3ex] \overline{DA} = \sqrt{100} \\[3ex] \overline{DA} = 10 $(259.) GCSE A circle, centre O, has a radius of 4cm. A and B are points on the circumference of the circle. Lines PA and PB are both tangents to the circle. PB = 12cm. (a) What is the length of PA? State the circle theorem you have used to find your answer. (b) What is the size of$P\hat{A}O$? State the circle theorem you have used to find your answer. (c) Calculate the area of the quadrilateral PAOB.$ (a) \\[3ex] |PA| = |PB| = 12\;cm ...two\;\;tangents:\;\;PA\;\;and\;\;PB\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] (b) \\[3ex] P\hat{A}O = 90^\circ ...radius\;OA \perp tangent\;PA\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] (c) \\[3ex] Quadrilateral\;PAOB = \triangle PAO + \triangle PBO \\[3ex] \underline{\triangle} \\[3ex] |OA| = |OB| = radius = perpendicular\;\;height = 4\;cm \\[3ex] |PB| = |PA| = base = 12\;cm \\[3ex] Area = \dfrac{1}{2} * base * perpendicular\;\;height \\[5ex] Area\;\;of\;\;Quadrilateral\;PAOB \\[3ex] = Area\;\;of\;\;\triangle PAO + Area\;\;of\;\;\triangle PBO \\[3ex] = \dfrac{1}{2} * 12 * 4 + \dfrac{1}{2} * 12 * 4 \\[5ex] = 24 + 24 \\[3ex] = 48\;cm^2 $(260.) ICSE In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC // AE. If ∠BAC = 50°, find giving reasons: (i) ∠ACB (ii) ∠EDC (iii) ∠BEC Hence prove that BE is also a diameter$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (i) \\[3ex] \angle ACB + \angle BAC + \angle ABC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle ACB + 50 + 90 = 180 \\[3ex] \angle ACB + 140 = 180 \\[3ex] \angle ACB = 180 - 140 \\[3ex] \angle ACB = 40^\circ \\[3ex] (ii) \\[3ex] \angle EAC = \angle ACB = 40^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{Quadrilateral\;\;AEDC} \\[3ex] \angle D + \angle A = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle D = \angle EDC ...diagram \\[3ex] \angle A = \angle EAC = 40^\circ ... diagram \\[3ex] \implies \\[3ex] \angle EDC + \angle EAC = 180 \\[3ex] \angle EDC + 40 = 180 \\[3ex] \angle EDC = 180 - 40 \\[3ex] \angle EDC = 140^\circ \\[3ex] (iii) \\[3ex] $Construction: Draw the chord from point B to point E and draw the chord from point E to point C Because ∠EAB = 90° (40° + 50°) chord BE is a diameter...angle in a semicircle This implies that chord BE (diameter) is perpendicular to chord AC (diameter) Let K be the midpoint of chord BE$ \angle KEC + \angle EKC + \angle ECK = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle KEC \\[3ex] \angle KEC + 90 + 50 = 180 \\[3ex] \angle KEC + 140 = 180 \\[3ex] \angle KEC = 180 - 140 \\[3ex] \angle KEC = 40^\circ \\[3ex] \angle BEC = \angle KEC = 40^\circ ... diagram $(261.) KCSE In the figure below, points A, B, C, D and E lie on the circumference of a circle centre O Line FAG is a tangent to the circle at A Chord DE of the circle is produced to intersect with the tangent at F Angle FAE = 30°, ∠EDC = 110° and ∠OCB = 55° (a) Determine the size of: (i) ∠AEC (ii) ∠AEB (b) Given that AB = 5 cm, ED = 4.4 cm and FE = 2.5 cm Calculate correct to 1 decimal place: (i) the radius of the circle (ii) the length of line AF$ (a) \\[3ex] (i) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; EBCD} \\[3ex] \angle B + \angle D = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle B + 110 = 180 \\[3ex] \angle B = 180 - 110 \\[3ex] \angle B = 70^\circ \\[3ex] \angle EBC = \angle B = 70^\circ ... diagram \\[3ex] \angle ABE = \angle FAE = 30^\circ ...\angle\;\;between\;\;tangent\;FAG\;\;and\;\;chord\;AE = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; ABCE} \\[3ex] \angle B = \angle ABE + \angle EBC ...diagram \\[3ex] \angle B = 30 + 70 \\[3ex] \angle B = 100^\circ \\[3ex] \angle E + \angle B = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle E + 100 = 180 \\[3ex] \angle E = 180 - 100 \\[3ex] \angle E = 80^\circ \\[3ex] \angle AEC = \angle E = 80^\circ ... diagram \\[3ex] (ii) \\[3ex] $Construction: Draw the radius from point O to point E$ Reflex\;\; \angle EOC = 2 * \angle EDC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\; \angle EOC = 2 * 110 \\[3ex] Reflex\;\; \angle EOC = 220^\circ \\[3ex] Obtuse\;\;\angle EOC + Reflex\;\;\angle EOC = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] Obtuse\;\;\angle EOC + 220 = 360 \\[3ex] Obtuse\;\;\angle EOC = 360 - 220 \\[3ex] Obtuse\;\;\angle EOC = 140^\circ \\[3ex] \underline{\triangle EOC} \\[3ex] \angle OEC = \angle OCE = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EOC \\[3ex] \angle OEC + \angle OCE + Obtuse\;\;\angle EOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOC \\[3ex] p + p + 140 = 180 \\[3ex] 2p = 180 - 140 \\[3ex] 2p = 40 \\[3ex] p = \dfrac{40}{2} \\[5ex] p = 20 \\[3ex] \therefore \angle OEC = \angle OCE = 20^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; ABCE} \\[3ex] \angle C = \angle OCB + \angle OCE ... diagram \\[3ex] \angle C = 55 + 20 \\[3ex] \angle C = 75^\circ \\[3ex] \angle A + \angle C = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle A + 75 = 180 \\[3ex] \angle A = 180 - 75 \\[3ex] \angle A = 105^\circ \\[3ex] \angle EAB = \angle A = 105^\circ ... diagram \\[3ex] \underline{\triangle AEB} \\[3ex] \angle AEB + \angle EAB + \angle ABE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABE \\[3ex] \angle AEB + 105 + 30 = 180 \\[3ex] \angle AEB + 135 = 180 \\[3ex] \angle AEB = 180 - 135 \\[3ex] \angle AEB = 45^\circ \\[3ex] (b) \\[3ex] (i) \\[3ex] \underline{\triangle EAB} \\[3ex] Let\;\;|EB| = k \\[3ex] \dfrac{|EB|}{\sin \angle EAB} = \dfrac{|AB|}{\sin \angle AEB} ...Sine\;\;Law \\[5ex] \dfrac{k}{\sin 105} = \dfrac{5}{\sin 45} \\[5ex] k = \dfrac{5\sin 105}{\sin 45} \\[5ex] \underline{\triangle EBC} \\[3ex] Let\;\;|EC| = m \\[3ex] \dfrac{|EC|}{\sin \angle EBC} = \dfrac{|EB|}{\sin \angle BCE} ...Sine\;\;Law \\[5ex] \angle BCE = \angle OCB + \angle OCE ... diagram \\[3ex] \angle BCE = 55 + 20 \\[3ex] \angle BCE = 75^\circ \\[3ex] \implies \\[3ex] \dfrac{m}{\sin 70} = \dfrac{k}{\sin 75} \\[5ex] \dfrac{m}{\sin 70} = k \div \sin 75 \\[5ex] \dfrac{m}{\sin 70} = \dfrac{5\sin 105}{\sin 45} \div \sin 75 \\[5ex] \dfrac{m}{\sin 70} = \dfrac{5\sin 105}{\sin 45} * \dfrac{1}{\sin 75} \\[5ex] m = \dfrac{5 * \sin 105 * \sin 70}{\sin 45 * \sin 75} \\[5ex] m = \dfrac{4.538366856}{0.6830127019} \\[5ex] m = 6.644630244 \\[3ex] \underline{\triangle EOC} \\[3ex] Radius = |OE| = |OC| = r \\[3ex] |EC|^2 = |OE|^2 + |OC|^2 - 2(|OE|)(|OC|) \cos \angle EOC...Cosine\;\;Law \\[3ex] m^2 = r^2 + r^2 - 2(r)(r) * \cos 140 \\[3ex] m^2 = 2r^2 - 2r^2 \cos 140 \\[3ex] m^2 = 2r^2(1 - \cos 140) \\[3ex] 2r^2(1 - \cos 140) = m^2 \\[3ex] 2 * r^2 * (1 - \cos 140) = m^2 \\[3ex] r^2 = \dfrac{m^2}{2(1 - \cos 140)} \\[5ex] r^2 = \dfrac{6.644630244^2}{2(1 - (-0.7660444431))} \\[5ex] r^2 = \dfrac{44.15111108}{2(1 + 0.7660444431)} \\[5ex] r^2 = \dfrac{44.15111108}{2(1.7660444431)} \\[5ex] r^2 = \dfrac{44.15111108}{3.532088886} \\[5ex] r^2 = 12.5 \\[3ex] r = \sqrt{12.5} \\[3ex] r = 3.535533906 \\[3ex] \therefore radius \approx 3.5\;cm \\[3ex] (ii) \\[3ex] $We can solve (ii) using at least two approaches. Use any approach you prefer. For KCSE students, use the 1st approach.$ \underline{1st\;\;Approach:\;\;Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] |AF|^2 = |FE| * |FD| \\[3ex] But:\;\;|FD| = |FE| + |ED| ... diagram \\[3ex] |AF|^2 = 2.5(|FE| + |ED|) \\[3ex] |AF|^2 = 2.5(2.5 + 4.4) \\[3ex] |AF|^2 = 2.5(6.9) \\[3ex] |AF|^2 = 17.25 \\[3ex] |AF| = \sqrt{17.25} \\[3ex] |AF| = 4.153311931 \\[3ex] |AF| \approx 4.2\;cm \\[3ex] \underline{2nd\;\;Approach:\;\;Sine\;\;Law} \\[3ex] \underline{\triangle EDC} \\[3ex] \dfrac{\sin \angle ECD}{|ED|} = \dfrac{\sin \angle EDC}{|EC|} ...Sine\;\;Law \\[5ex] \dfrac{\sin \angle ECD}{4.4} = \dfrac{\sin 110}{m} \\[5ex] \dfrac{\sin \angle ECD}{4.4} = \dfrac{\sin 110}{6.644630244} \\[5ex] \sin \angle ECD = \dfrac{4.4 \sin 110}{6.644630244} \\[5ex] \sin \angle ECD = \dfrac{4.134647531}{6.644630244} \\[5ex] \sin \angle ECD = 0.6222539674 \\[3ex] \angle ECD = \sin^{-1}(0.6222539674) \\[3ex] \angle ECD = 38.48091859 \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; EBCD} \\[3ex] \angle C = \angle OCB + \angle OCE + \angle ECD ...diagram \\[3ex] \angle C = 55 + 20 + 38.48091859 \\[3ex] \angle C = 113.4809186^\circ \\[3ex] \angle BEF = \angle C = 113.4809186^\circ ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] But: \\[3ex] \angle AEF + \angle AEB = \angle BEF ... diagram \\[3ex] \angle AEF + 45 = 113.4809186 \\[3ex] \angle AEF = 113.4809186 - 45 \\[3ex] \angle AEF = 68.48091859^\circ \\[3ex] \underline{\triangle AEF} \\[3ex] Let\;\;|AF| = n \\[3ex] \dfrac{|AF|}{\sin \angle AEF} = \dfrac{|FE|}{\sin \angle FAE} ...Sine\;\;Law \\[5ex] \dfrac{n}{\sin 68.48091859} = \dfrac{2.5}{\sin 30} \\[5ex] n = \dfrac{2.5 * \sin 68.48091859}{\sin 30} \\[5ex] n = \dfrac{2.325738648}{0.5} \\[5ex] n = 4.651477296 \\[3ex] \therefore |AF| \approx 4.7\;cm $(262.) GCSE AOC and BOD are diameters of a circle, centre O Prove that triangle ABC and triangle DCB are congruent. (a) There are at least two ways to prove that$\triangle ABC$and$\triangle DCB$are congruent Use any approach you prefer$ \underline{First\;\;Approach:\;\;HL} \\[3ex] \angle ABC = \angle DCB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \therefore \triangle ABC \;\;and\;\;\triangle DBC \;\;are\;\;right\;\;\triangle s \\[3ex] \implies \\[3ex] |AC| = |DB| = diameter = hypotenuse \;(H) \\[3ex] |BC| = |CB| ...diagram = leg\;(L) \\[3ex] \implies \\[3ex] HL\;\;Congruency \\[5ex] \underline{Second\;\;Approach:\;\;AAS} \\[3ex] \angle ABC = \angle DCB = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] ...angle\;(A) \\[3ex] \angle BAC = \angle CDB ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] ...angle\;(A) \\[3ex] |AC| = |DB| = diameter \\[3ex] ..side\;(S) \\[3ex] \implies \\[3ex] AAS\;\;Congruency $WASSCE The diagram shows a circle centre O Use it to answer questions 263 and 264 (263.) Find the value of y$ A.\;\; 43^\circ \\[3ex] B.\;\; 47^\circ \\[3ex] C.\;\; 54^\circ \\[3ex] D.\;\; 89^\circ \\[3ex] $Construction: Label the points on the circle$ \angle OBC = 43^\circ ... vertical\;\;\angle s\;\;are\;\;equal \\[3ex] y = \angle OBC = \angle OCB = 43^\circ ...base\;\;\angle s \;\;of\;\;isosceles\;\;\triangle OBC $(264.) Find the value x$ A.\;\; 43^\circ \\[3ex] B.\;\; 47^\circ \\[3ex] C.\;\; 54^\circ \\[3ex] D.\;\; 89^\circ \\[3ex] $Construction: Label the points on the circle$ \angle BOC + \angle OBC + \angle OCB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OBC \\[3ex] \angle BOC + 43 + 43 = 180 \\[3ex] \angle BOC + 86 = 180 \\[3ex] \angle BOC = 180 - 86 \\[3ex] \angle BOC = 94^\circ \\[3ex] \angle BOC = 2 * x ...\angle\;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 94 = 2x \\[3ex] 2x = 94 \\[3ex] x = \dfrac{94}{2} \\[5ex] x = 47^\circ $(265.) GCSE A, B, C and D are points on a circle. TDV is the tangent to the circle at D. AB = AD Angle ADT = 71° Work out the size of angle BCD Give a reason for each stage of your working. Construction: Join the chord from Point D to Point B$ \angle ABD = 71^\circ ...\angle\;\;between\;\;tangent\;TDV\;\;and\;\;chord\;AD = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle ADB = \angle ABD = 71^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ABD \\[3ex] \angle DAB + \angle ADB + \angle ABD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \implies \\[3ex] \angle DAB + 71 + 71 = 180 \\[3ex] \angle DAB + 142 = 180 \\[3ex] \angle DAB = 180 - 142 \\[3ex] \angle DAB = 38^\circ \\[3ex] \angle BCD + \angle DAB = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BCD + 38 = 180 \\[3ex] \angle BCD = 180 - 38 \\[3ex] \angle BCD = 142^\circ $(266.) ACT Inscribed in the circle below is regular hexagon ABCDEF with some diagonals shown. What is the measure of ∠CAD?$ F.\;\; 27^\circ \\[3ex] G.\;\; 30^\circ \\[3ex] H.\;\; 37.5^\circ \\[3ex] J.\;\; 40^\circ \\[3ex] K.\;\; 60^\circ \\[3ex]  |AD|\;\;is\;\;a\;\;diameter \\[3ex] \implies \\[3ex] \angle ACD = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;polygon \\[3ex] = 180(n - 2) \\[3ex] For\;\;a\;\;regular\;\;hexagon:\;\;n = 6 \;\;(6\;\;sides) \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = 180(6 - 2) \\[3ex] = 180(4) \\[3ex] = 720^\circ \\[3ex] Each\;\;interior\;\;\angle\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = \dfrac{Sum}{n} \\[5ex] = \dfrac{720}{6} \\[5ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle A = \angle B = \angle C = \angle D = \angle E = \angle F = 120^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;ABCD} \\[3ex] \angle ABC = \angle B = 120^\circ ...diagram \\[3ex] \angle ABC + \angle CDA = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 120 + \angle CDA = 180 \\[3ex] \angle CDA = 180 - 120 \\[3ex] \angle CDA = 60^\circ \\[3ex] \underline{\triangle ACD} \\[3ex] \angle CAD + \angle CDA + \angle ACD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] \angle CAD + 60 + 90 = 180 \\[3ex] \angle CAD + 150 = 180 \\[3ex] \angle CAD = 180 - 150 \\[3ex] \angle CAD = 30^\circ $WASSCE Use the diagram to answer questions 267 and 268 (267.) Find the value of m$ A.\;\; 35^\circ \\[3ex] B.\;\; 45^\circ \\[3ex] C.\;\; 65^\circ \\[3ex] D.\;\; 75^\circ \\[3ex] $Construction: Label the points on the circle$ \angle CBD = \angle CDB = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle CBD \\[3ex] m = \angle CBD = 35^\circ ... \angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $(268.) Find the value of n$ A.\;\; 70^\circ \\[3ex] B.\;\; 65^\circ \\[3ex] C.\;\; 55^\circ \\[3ex] D.\;\; 35^\circ \\[3ex] $Construction: Label the points on the circle$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ABC = n + 35 ...diagram \\[3ex] \implies \\[3ex] n + 35 = 90 \\[3ex] n = 90 - 35 \\[3ex] n = 55^\circ $(269.) CSEC H, J, K, L and M are points on the circumference of a circle with centre O MK is a diameter of the circle and it is parallel to HJ MJ = JL and angle JMK = 38° (i) Explain, giving a reason, why angle (a) HJM = 38° (b) MJK = 90° (ii) Determine the value of EACH of the following angles. Show detailed working where appropriate. (a) Angle MLJ (b) Angle LJK (c) Angle JHM$ (i) \\[3ex] (a) \\[3ex] \angle HJM = 38^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|HJ| \;\;and\;\; |MK|\;\;are\;\;equal \\[3ex] (b) \\[3ex] \angle MJK = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] (ii) \\[3ex] (a) \\[3ex] \underline{\triangle MJK} \\[3ex] \angle MKJ + \angle MJK + \angle JMK = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle MKJ + 90 + 38 = 180 \\[3ex] \angle MKJ + 128 = 180 \\[3ex] \angle MKJ = 180 - 128 \\[3ex] \angle MKJ = 52^\circ \\[3ex] \angle MLJ = \angle MKJ = 52^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (b) \\[3ex] $We can solve (b) using at least two appraoches. Use any approach you prefer.$ \underline{First\;\;Approach} \\[3ex] \underline{\triangle MJL} \\[3ex] \angle LMJ = \angle MLJ = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle MJL \\[3ex] But:\;\; \angle LMK + \angle JMK = \angle LMJ ... diagram \\[3ex] \angle LMK + 38 = 52 \\[3ex] \angle LMK = 52 - 38 \\[3ex] \angle LMK = 14^\circ \\[3ex] \angle LJK = \angle LMK = 14^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{Second\;\;Approach} \\[3ex] \underline{\triangle MJL} \\[3ex] \angle LMJ = \angle MLJ = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle MJL \\[3ex] \angle MJL + \angle LMJ + \angle MLJ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle MJL + 52 + 52 = 180 \\[3ex] \angle MJL + 104 = 180 \\[3ex] \angle MJL = 180 - 104 \\[3ex] \angle MJL = 76^\circ \\[3ex] \angle MJL + \angle LJK = \angle MJK ...diagram \\[3ex] 76 + \angle LJK = 90 \\[3ex] \angle LJK = 90 - 76 \\[3ex] \angle LJK = 14^\circ \\[3ex] (c) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;LMHJ} \\[3ex] \angle H + \angle L = 180^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;LMHJ \\[3ex] \angle L = \angle MLJ = 52^\circ ... diagram \\[3ex] \angle H = \angle JHM ... diagram \\[3ex] \implies \\[3ex] \angle JHM + 52 = 180 \\[3ex] \angle JHM = 180 - 52 \\[3ex] \angle JHM = 128^\circ $(270.) GCSE A, B, C and D are points on a circle, centre O. AOC is a diameter of the circle. Angle AOD = 98° Work out the size of angle DBC Give a reason for each stage in your working.$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle AOD = 2 * \angle ABD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 98 = 2 * \angle ABD \\[3ex] 2 * \angle ABD = 98 \\[3ex] \angle ABD = \dfrac{98}{2} \\[5ex] \angle ABD = 49^\circ \\[3ex] \angle ABC = \angle ABD + \angle DBC ...diagram \\[3ex] 90 = 49 + \angle DBC \\[3ex] 49 + \angle DBC = 90 \\[3ex] \angle DBC = 90 - 49 \\[3ex] \angle DBC = 41^\circ $WASSCE In the diagram, O is the centre of the circle QRST ∠QRT = 42° and ∠PQS = 124° Use it to answer questions 271 and 272 (271.) Find the size of ∠RSQ$ A.\;\; 34^\circ \\[3ex] B.\;\; 48^\circ \\[3ex] C.\;\; 56^\circ \\[3ex] D.\;\; 76^\circ \\[3ex]  \angle QST = \angle QRT = 42^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle RST = 90^\circ ... \angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle RSQ + \angle QST = \angle RST ...diagram \\[3ex] \angle RSQ + 42 = 90 \\[3ex] \angle RSQ = 90 - 42 \\[3ex] \angle RSQ = 48^\circ $(272.) Find ∠STR$ A.\;\; 76^\circ \\[3ex] B.\;\; 56^\circ \\[3ex] C.\;\; 34^\circ \\[3ex] D.\;\; 21^\circ \\[3ex]  \angle SQR + \angle PQS = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle SQR + 124 = 180 \\[3ex] \angle SQR = 180 - 124 \\[3ex] \angle SQR = 56^\circ \\[3ex] \angle STR = \angle SQR = 56^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $(273.) GCSE In the diagram, A, B, C and D are points on a circle, centre OCBO = 62° and ∠BCD = 126° (a) Find the size, in degrees, of ∠BAD Give a reason for your answer. (b) Find the size, in degrees, of ∠ODC Give reasons for your working.$ (a) \\[3ex] \angle BAD + \angle BCD = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BAD + 126 = 180 \\[3ex] \angle BAD = 180 - 126 \\[3ex] \angle BAD = 54^\circ \\[5ex] (b) \\[3ex] Reflex\;\;\angle BOD = 2 * \angle BCD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle BOD = 2 * 126 \\[3ex] Reflex\;\;\angle BOD = 252^\circ \\[3ex] Obtuse\;\;\angle BOD + Reflex\;\;\angle BOD = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] Obtuse\;\;\angle BOD + 252 = 360 \\[3ex] Obtuse\;\;\angle BOD = 360 - 252 \\[3ex] Obtuse\;\;\angle BOD = 108^\circ \\[3ex] \angle ODC + \angle BCD + \angle CBO + Obtuse\;\;\angle BOD = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;CBOD \\[3ex] \angle ODC + 126 + 62 + 108 = 360 \\[3ex] \angle ODC + 296 = 360 \\[3ex] \angle ODC = 360 - 296 \\[3ex] \angle ODC = 64^\circ $(274.) ACT Regular hexagon ABCDEF is inscribed in a circle, as shown below. If the length of radius$\overline{OD}$is 15 centimeters, how long is$\overline{AB}$, in centimeters?$ A.\;\; 15 \\[3ex] B.\;\; 18 \\[3ex] C.\;\; 30 \\[3ex] D.\;\; 5\pi \\[3ex] E.\;\; \dfrac{225\pi}{6} \\[5ex] $Construction: Join the chord from Point B to Point E$ Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;polygon \\[3ex] = 180(n - 2) \\[3ex] For\;\;a\;\;regular\;\;hexagon:\;\;n = 6 \;\;(6\;\;sides) \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = 180(6 - 2) \\[3ex] = 180(4) \\[3ex] = 720^\circ \\[3ex] Each\;\;interior\;\;\angle\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = \dfrac{Sum}{n} \\[5ex] = \dfrac{720}{6} \\[5ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle A = \angle B = \angle C = \angle D = \angle E = \angle F = 120^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;EBCD} \\[3ex] \angle BCD = \angle C = 120^\circ ...diagram \\[3ex] \angle BED + \angle BCD = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle BED + 120 = 180 \\[3ex] \angle BED = 180 - 120 \\[3ex] \angle BED = 60^\circ \\[3ex] \underline{\triangle OED} \\[3ex] \angle OED = \angle BED = 60^\circ ... diagram \\[3ex] \angle OED = \angle ODE = 60^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OED \\[3ex] \implies \\[3ex] |OE| = |OD| = 15\;cm ... isosceles\;\; \triangle OED \\[3ex] \angle EOD + \angle OED + \angle ODE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OED \\[3ex] \angle EOD + 60 + 60 = 180 \\[3ex] \angle EOD + 120 = 180 \\[3ex] \angle EOD = 180 - 120 \\[3ex] \angle EOD = 60^\circ \\[3ex] \implies \\[3ex] \triangle OED \;\;is\;\;an\;\;equilateral\;\;\triangle \\[3ex] \angle EOD = \angle OED = \angle ODE = 60^\circ \\[3ex] Similarly: \\[3ex] |OD| = |OE| = |ED| = 15\;cm ...equilateral\;\; \triangle OED \\[3ex] \implies \\[3ex] |AB| = |ED| = 15\;cm ...regular\;\;hexagon $WASSCE In the diagram, RT is a tangent to the circle at R, ∠PQR = 70°, ∠QRT = 52°, ∠QSR = y and ∠PRQ = x Use the diagram to answer questions 275 and 276 (275.) Find the value of y$ A.\;\; 18^\circ \\[3ex] B.\;\; 52^\circ \\[3ex] C.\;\; 60^\circ \\[3ex] D.\;\; 70^\circ \\[3ex]  \angle RPQ = \angle QSR = y ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle QRT = \angle RPQ ...\angle\;\;between\;\;tangent\;RT \;\;and\;\;chord\;RQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] 52 = y \\[3ex] y = 52^\circ $(276.) Calculate the value of x$ A.\;\; 48^\circ \\[3ex] B.\;\; 55^\circ \\[3ex] C.\;\; 58^\circ \\[3ex] D.\;\; 70^\circ \\[3ex]  \underline{\triangle RPQ} \\[3ex] \angle PRQ + \angle RPQ + \angle PQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] x + y + 70 = 180 \\[3ex] x + 52 + 70 = 180 \\[3ex] x + 122 = 180 \\[3ex] x = 180 - 122 \\[3ex] x = 58^\circ $(277.) GCSE In Figure 3, ABCD is a circle, centre O EA is the tangent to the circle at A ECF is the tangent to the circle at C EDO is a straight line. OEC = 27° ∠BCF = 59° ∠EC = 12cm (a) Explain why ∠OCE = 90° (b) Calculate the area, in$cm^2$to 3 significant figures, of$\triangle$OEC (c) Giving reasons, calculate the size, in degrees, of ∠ADC (d) Calculate the size, in degrees, of ∠ADC (e) Calculate the size, in degrees, of ∠BAO$ (a) \\[3ex] \angle OCE = 90^\circ ...radius\;OC \perp tangent\;ECF\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] (b) \\[3ex] \underline{\triangle OEC} \\[3ex] \triangle OEC \;\;is\;\;a\;\;right\;\;\triangle \\[3ex] SOHCAHTOA \\[3ex] |OC| = radius = r \\[3ex] \tan 27^\circ = \dfrac{opp}{adj} = \dfrac{|OC|}{|EC|} \\[5ex] \tan 27 = \dfrac{r}{12} \\[5ex] r = |OC| = 12\tan 27 \\[3ex] Area = \dfrac{|EC| * |OC|}{2} \\[5ex] Area = \dfrac{12 * 12 * \tan 27}{2} \\[5ex] Area = 72\tan 27 \\[3ex] Area = 72(0.5095254495) \\[3ex] Area = 36.68583236 \\[3ex] Area \approx 36.7cm^2 ...to\;\;3\;\;significant\;\;figures \\[3ex] (c) \\[3ex] \angle EOC + \angle OEC + \angle OCE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OEC \\[3ex] \angle EOC + 27 + 90 = 180 \\[3ex] \angle EOC + 117 = 180 \\[3ex] \angle EOC = 180 - 117 \\[3ex] \angle EOC = 63^\circ \\[3ex] Similarly: \\[3ex] \angle OAE = 90^\circ ...radius\;OA \perp tangent\;EA\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OEA = \angle OEC = 27^\circ \\[3ex] ... two\;\;tangents:\;AE\;\;and\;\;CE\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;E\;\;bisects\;\;the\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents:\;\angle AEC \\[3ex] \implies \\[3ex] \angle EOA = 63^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OEA \\[3ex] Obtuse\;\;\angle AOC = \angle EOA + \angle EOC ...diagram \\[3ex] Obtuse\;\;\angle AOC = 63 + 63 \\[3ex] Obtuse\;\;\angle AOC = 126^\circ \\[3ex] Obtuse\;\;\angle AOC = 2 * \angle ABC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 126 = 2 * \angle ABC \\[3ex] 2 * \angle ABC = 126 \\[3ex] \angle ABC = \dfrac{126}{2} \\[5ex] \angle ABC = 63^\circ \\[3ex] (d) \\[3ex] Obtuse\;\;\angle AOC + Reflex\;\;\angle AOC = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] 126 + Reflex\;\;\angle AOC = 360 \\[3ex] Reflex\;\;\angle AOC = 360 - 126 \\[3ex] Reflex\;\;\angle AOC = 234^\circ \\[3ex] Reflex\;\;\angle AOC = 2 * \angle ADC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 234 = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 234 \\[3ex] \angle ADC = \dfrac{234}{2} \\[5ex] \angle ADC = 117^\circ \\[3ex] $(e) Let us do some construction to see the theorem we need to apply (Alternate Segment Theorem) before we find ∠BAO$ \angle BAC = \angle BCF = 59^\circ ...\angle\;\;between\;\;tangent\;BCF\;\;and\;\;chord\;BC = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle AOC} \\[3ex] \angle OAC = \angle OCA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOC \\[3ex] \angle OAC + \angle OCA + Obtuse\;\;\angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] p + p + 126 = 180 \\[3ex] 2p = 180 - 126 \\[3ex] 2p = 54 \\[3ex] p = \dfrac{54}{2} \\[5ex] p = 27 \\[3ex] \therefore \angle OAC = \angle OCA = 27^\circ \\[3ex] \angle BAC = \angle BAO + \angle OAC ...diagram \\[3ex] 59 = \angle BAO + 27 \\[3ex] \angle BAO + 27 = 59 \\[3ex] \angle BAO = 59 - 27 \\[3ex] \angle BAO = 32^\circ $(278.) ICSE PQRS is a cyclic quadrilateral. Given ∠QPS = 73° ∠PQS = 55° and ∠PSR = 82°, calculate: (i) ∠QRS (ii) ∠RQS (iii) ∠PRQ$ (i) \\[3ex] \angle QRS + \angle QPS = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QRS + 73 = 180 \\[3ex] \angle QRS = 180 - 73 \\[3ex] \angle QRS = 107^\circ \\[3ex] (ii) \\[3ex] $We can solve (ii) using at least two approaches. Use any approach that you prefer.$ \underline{First\;\;Approach} \\[3ex] \angle PQR + \angle PSR = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle PQR + 82 = 180 \\[3ex] \angle PQR = 180 - 82 \\[3ex] \angle PQR = 98^\circ \\[3ex] But:\;\; \angle RQS + \angle PQS = \angle PQR ... diagram \\[3ex] \angle RQS + 55 = 98 \\[3ex] \angle RQS = 98 - 55 \\[3ex] \angle RQS = 43^\circ \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle PSQ + \angle QPS + \angle PQS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] \angle PSQ + 73 + 55 = 180 \\[3ex] \angle PSQ + 128 = 180 \\[3ex] \angle PSQ = 180 - 128 \\[3ex] \angle PSQ = 52^\circ \\[3ex] But:\;\;\angle PSQ + \angle QSR = \angle PSR ...diagram \\[3ex] 52 + \angle QSR = 82 \\[3ex] \angle QSR = 82 - 52 \\[3ex] \angle QSR = 30^\circ \\[3ex] \angle RQS + \angle QSR + \angle QRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] \angle RQS + 30 + 107 = 180 \\[3ex] \angle RQS + 137 = 180 \\[3ex] \angle RQS = 180 - 137 \\[3ex] \angle RQS = 43^\circ \\[3ex] (iii) \\[3ex] $Construction: Draw the chord from point P to point R Use a red color$ \angle PRQ = \angle PSQ = 52^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] $(279.) ICSE In the figure given below 'O' is the centre of the circle. If QR = OP and ∠ORP = 20°, find the value of 'x' giving reasons.$ \angle ORP = \angle ORQ = \angle QRO = 20^\circ ... diagram \\[3ex] |OP| = |OQ| ...radius \\[3ex] \angle OPQ = \angle OQP ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OPQ \\[3ex] |QR| = |OP| ... given \\[3ex] \implies \\[3ex] |QR| = |OQ| ...transitive\;\;property \\[3ex] \implies \\[3ex] \angle QOR = \angle QRO = 20^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle QOR \\[3ex] \angle OQP = \angle QOR + \angle QRO ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle OQP = 20 + 20 \\[3ex] \angle OQP = 40^\circ \\[3ex] \angle OPQ = \angle OQP = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OPQ \\[3ex] \angle POQ + \angle OPQ + \angle OQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPQ \\[3ex] \angle POQ + 40 + 40 = 180 \\[3ex] \angle POQ + 80 = 180 \\[3ex] \angle POQ = 180 - 80 \\[3ex] \angle POQ = 100^\circ \\[3ex] x + \angle POQ + \angle QOR = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] x + 100 + 20 = 180 \\[3ex] x + 120 = 180 \\[3ex] x = 180 - 120 \\[3ex] x = 60^\circ $(280.) ACT The circle shown below has a radius of 10 meters, and the length of chord$\overline{AB}$is 16 meters. If O marks the center of the circle, what is the length of$\overline{OC}$?$ A.\;\; 2\sqrt{3} \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 12 \\[3ex] D.\;\; 4\sqrt{21} \\[3ex] E.\;\; 36 \\[3ex] $Construction: Draw the radius from Point O to POint A$ |AC| = |CB| = \dfrac{|AB|}{2} \\[5ex] ...line\;OA\;\;\perp\;\;to\;\;chord\;AB;\;\;bisects\;\;the\;\;chord \\[3ex] |AC| = \dfrac{16}{2} = 8\;cm \\[5ex] |OA| = 10\;cm ...radius \\[3ex] \underline{Right\;\triangle\;\;OCA} \\[3ex] |OC|^2 + |AC|^2 = |OA|^2 ...Pythagorean\;\;theorem \\[3ex] |OC|^2 + 8^2 = 10^2 \\[3ex] |OC|^2 + 64 = 100 \\[3ex] |OC|^2 = 100 - 64 \\[3ex] |OC|^2 = 36 \\[3ex] |OC| = \sqrt{36} \\[3ex] |OC| = 6\;cm $(281.) KCSE In the figure below A, B, C and D are points on the circumference of the circle, centre O. A tangent to the circle at A intersects chord CD produced at E. Line AB is parallel to line EC. Angle AED = 45° and angle ABD = 40°. (a) Calculate the size of: (i) ∠ADB (ii) ∠OCD (b) Given that ED = 3.5 cm and DC = 4.9 cm, calculate correct to 1 decimal place: (i) the length of the tangent AE (ii) the radius of the circle.$ (a) \\[3ex] \angle CDB = 40^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|EC| \;\;and\;\; |AB|\;\;are\;\;equal \\[3ex] \angle DAE = 40^\circ ...\angle\;\;between\;\;tangent\;EA\;\;and\;\;chord\;DA = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle EDA + \angle DAE + \angle AED = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AED \\[3ex] \angle EDA + 40 + 45 = 180 \\[3ex] \angle EDA + 85 = 180 \\[3ex] \angle EDA = 180 - 85 \\[3ex] \angle EDA = 95^\circ \\[3ex] (i) \\[3ex] \angle EDA + \angle ADB + \angle CDB = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 95 + \angle ADB + 40 = 180 \\[3ex] \angle ADB + 135 = 180 \\[3ex] \angle ADB = 180 - 135 \\[3ex] \angle ADB = 45^\circ \\[3ex] (ii) \\[3ex] $Construction: Draw the chord from point B to point C This forms the cyclic quadrilateral ABCD$ Obtuse\;\angle BOC = 2 * \angle CDB ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * 40 \\[3ex] \angle BOC = 80^\circ \\[3ex] \underline{\triangle BOC} \\[3ex] \angle OCB = \angle OBC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle BOC \\[3ex] \angle OCB + \angle OBC + \angle BOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] p + p + 80 = 180 \\[3ex] 2p = 180 - 80 \\[3ex] 2p = 100 \\[3ex] p = \dfrac{100}{2} \\[5ex] p = 50 \\[3ex] \therefore \angle OCB = \angle OBC = 50^\circ \\[3ex] \underline{\triangle ADB} \\[3ex] \angle ADB + \angle DAB + \angle ABD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] 45 + \angle DAB + 40 = 180 \\[3ex] \angle DAB + 85 = 180 \\[3ex] \angle DAB = 180 - 85 \\[3ex] \angle DAB = 95^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle A + \angle C = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle A = \angle DAB = 95^\circ \\[3ex] \angle C = \angle OCD + \angle OCB ... diagram \\[3ex] \angle C = \angle OCD + 50 \\[3ex] \implies \\[3ex] 95 + \angle OCD + 50 = 180 \\[3ex] \angle OCD + 145 = 180 \\[3ex] \angle OCD = 180 - 145 \\[3ex] \angle OCD = 35^\circ \\[3ex] (b) \\[3ex] $Label: the lengths Let the radius = r We can solve (i) using at least two approaches. Use any approach you prefer. For KCSE students, use the 1st approach.$ \underline{1st\;\;Approach:\;\;Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] |AE|^2 = |ED| * |EC| \\[3ex] But:\;\; |EC| = |ED| + |DC| ...diagram \\[3ex] \implies \\[3ex] |AE|^2 = 3.5(3.5 + 4.9) \\[3ex] |AE|^2 = 3.5(8.4) \\[3ex] |AE|^2 = 29.4 \\[3ex] |AE| = \sqrt{29.4} \\[3ex] |AE| = 5.422176685 \\[3ex] |AE| \approx 5.4\;cm \\[3ex] \underline{2nd\;\;Approach:\;\;Sine\;\;Law} \\[3ex] \underline{\triangle AED} \\[3ex] \dfrac{|AE|}{\sin \angle EDA} = \dfrac{|ED|}{\sin \angle DAE} ... Sine\;\;Law \\[5ex] \dfrac{|AE|}{\sin 95} = \dfrac{3.5}{\sin 40} \\[5ex] |AE| = \dfrac{3.5 * \sin 95}{\sin 40} \\[5ex] |AE| = \dfrac{3.486681443}{0.6427876097} \\[5ex] |AE| = 5.424313398 \\[3ex] |AE| \approx 5.4\;cm \\[3ex] (ii) \\[3ex] \underline{\triangle BDC} \\[3ex] \angle DBC + \angle BCD + \angle CDB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle BCD = \angle OCB + \angle OCD ... diagram \\[3ex] \angle BCD = 50 + 35 \\[3ex] \angle BCD = 85^\circ \\[3ex] \implies \\[3ex] \angle DBC + 85 + 40 = 180 \\[3ex] \angle DBC + 125 = 180 \\[3ex] \angle DBC = 180 - 125 \\[3ex] \angle DBC = 55^\circ \\[3ex] \dfrac{|BC|}{\sin \angle CDB} = \dfrac{|DC|}{\sin \angle DBC} ... Sine\;\;Law \\[5ex] \dfrac{|BC|}{\sin 40} = \dfrac{4.9}{\sin 55} \\[5ex] |BC| = \dfrac{4.9 * \sin 40}{\sin 55} ... eqn.(1) \\[5ex] \underline{\triangle BOC} \\[3ex] \dfrac{|BC|}{\sin \angle BOC} = \dfrac{|OC|}{\sin \angle OBC} ... Sine\;\;Law \\[5ex] |OC| = radius = r \\[3ex] \implies \\[3ex] \dfrac{|BC|}{\sin 80} = \dfrac{r}{\sin 50} \\[5ex] |BC| = \dfrac{r * \sin 80}{\sin 50} ... eqn.(2) \\[5ex] |BC| = |BC| \implies eqn.(1) = eqn.(2) \\[3ex] \dfrac{r \sin 80}{\sin 50} = \dfrac{4.9\sin 40}{\sin 55} \\[5ex] r = \dfrac{4.9 * \sin 40 * \sin 50}{\sin 80 * \sin 55} \\[5ex] r = \dfrac{2.412778995}{0.8067072841} \\[5ex] r = 2.990897743 \\[3ex] r \approx 3.0\;cm $(282.) NYSED Regents Examination In the diagram below of circle O, chords$\overline{JT}$and$\overline{ER}$intersect at M If EM = 8 and RM = 15, the lengths of$\overline{JM}$and$\overline{TM}$could be (1) 12 and 9.5 (3) 16 and 7.5 (2) 14 and 8.5 (4) 18 and 6.5$ JM * TM = EM * RM ...Intersecting\;\;Chords\;\;Theorem \\[3ex] JM * TM = 8 * 15 \\[3ex] JM * TM = 120 \\[3ex] $Test the options 12 * 9.5 = 114 14 * 8.5 = 119 16 * 7.5 = 120 18 * 6.5 = 117$\overline{JM}$and$\overline{TM}$could be 16 and 7.5 because the product gives 120 (283.) JAMB In the figure above, PT is a tangent to the circle with centre O. If$PQT = 30^\circ$, find the value of PTO$ A.\:\: 30^\circ \\[3ex] B.\:\: 15^\circ \\[3ex] C.\:\: 24^\circ \\[3ex] D.\:\: 12^\circ \\[3ex] E.\:\: 60^\circ \\[3ex]  \angle OPT = 90^\circ ... radius\;OP \perp tangent\;PT\:\:at\:\:point\:\:of\:\:contact\;P \\[3ex] \underline{\triangle QPT} \\[3ex] 30 + (x + 90) + (2x + x) = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 30 + x + 90 + 2x + x = 180 \\[3ex] 4x + 120 = 180 \\[3ex] 4x = 180 - 120 \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15^\circ \\[3ex] \angle PTO = 2x \\[3ex] \angle PTO = 2(15) \\[3ex] \angle PTO = 30^\circ $(284.) GCSE R and S are points on a circle with centre O (a) On the diagram above, shade a segment of the circle. (b) Write down the mathematical name of the straight line RS In the diagram below, P and Q are points on a circle with centre O QT is a tangent to the circle. Angle OPQ = 18° (c) Work out the size of angle PQT Give a reason for each stage of your working. (a) This is the minor segment (shaded blue) OR This is the major segment (shaded yellow) (b) The straight line RS is a chord.$ \angle OQP = \angle OPQ = 18^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OPQ \\[3ex] \angle OQT = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;QT\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle OQT = \angle OQP + \angle PQT ...diagram \\[3ex] 90 = 18 + \angle PQT \\[3ex] 18 + \angle PQT = 90 \\[3ex] \angle PQT = 90 - 18 \\[3ex] \angle PQT = 72^\circ $(285.) KCSE In the figure below, PQRS is a cyclic quadrilateral. PQ = QR, ∠PQR = 105° and PS is parallel to QR. Determine the size of: (a) ∠PSR (b) ∠PQS For part (a), we can solve it using at least two approaches Use any approach you prefer$ (a) \\[3ex] \underline{First\;\;Approach} \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;PQRS} \\[3ex] \angle PSR + \angle PQR = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle PSR + 105 = 180 \\[3ex] \angle PSR = 180 - 105 \\[3ex] \angle PSR = 75^\circ \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle QPR = \angle QRP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle PQR \\[3ex] \angle QPR + \angle QRP + \angle PQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] k + k + 105 = 180 \\[3ex] 2k = 180 - 105 \\[3ex] 2k = 75 \\[3ex] k = \dfrac{75}{2} \\[5ex] k = 37.5 \\[3ex] \therefore \angle QPR = \angle QRP = 37.5^\circ \\[3ex] \angle PSQ = \angle QRP = 37.5^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle QSR = \angle QPR = 37.5^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle PSR = \angle PSQ + \angle QSR ... diagram \\[3ex] \angle PSR = 37.5 + 37.5 \\[3ex] \angle PSR = 75^\circ \\[3ex] (b) \\[3ex] \angle SQR = \angle PSQ = 37.5^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|QR| \;\;and\;\; |PS|\;\;are\;\;equal \\[3ex] \angle PQS + \angle SQR = 105^\circ ... diagram \\[3ex] \angle PQS + 37.5 = 105 \\[3ex] \angle PQS = 105 - 37.5 \\[3ex] \angle PQS = 67.5^\circ $(286.) NYSED Regents Examination In a circle A below, chord$\overline{BC}$and diameter$\overline{DAE}$intersect at F If m$\overset{\huge\frown}{CD}$= 46° and m$\overset{\huge\frown}{DB}$= 102°, what is m∠CFE?$ \angle CFE = \dfrac{\overset{\huge\frown}{DB} + \overset{\huge\frown}{CE}}{2} \\[5ex] ... Angle\;\;of\;\;Intersecting\;\;Chords\;\;Theorem \\[3ex] \overset{\huge\frown}{CD} + \overset{\huge\frown}{CE} = 180^\circ ...arcs\;\;of\;\;semicircle \\[3ex] \overset{\huge\frown}{CE} = 180 - \overset{\huge\frown}{CD} \\[3ex] \overset{\huge\frown}{CE} = 180 - 46 \\[3ex] \overset{\huge\frown}{CE} = 134^\circ \\[3ex] \implies \\[3ex] \angle CFE = \dfrac{102 + 134}{2} \\[5ex] \angle CFE = \dfrac{236}{2} \\[5ex] \angle CFE = 118^\circ $(287.) GCSE A, B and C are points on a circle. CD is a tangent. (a) Assume that triangle ABC is isosceles with AC = BC Prove that AB is parallel to DC (b) In fact, triangle ABC is equilateral. Tick the two boxes for the statements that must be correct.$ (a) \\[3ex] \angle CAB = \angle CBA ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ABC \;\;because\;\; AC = BC \\[3ex] \angle ACD = \angle CBA ...\angle\;\;between\;\;tangent\;CD\;\;and\;\;chord\;AC = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \implies \\[3ex] \angle CAB = \angle ACD...equality\;\;with\;\;\angle CBA \\[3ex] \implies \\[3ex] \angle CAB = \angle ACD ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|AB| \;\;and\;\; |DC|\;\;are\;\;equal \\[3ex] \therefore |AB| || |DC| \\[3ex] (b) \\[3ex] \angle CAB = \angle CBA = \angle ACB = 60^\circ ... \triangle ABC \;\;is\;\;equilateral\;\triangle \\[3ex] \underline{Options} \\[3ex] |AB| || |DC|...Yes\;\;because\;\;alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|AB| \;\;and\;\; |DC|\;\;are\;\;equal \\[3ex] |AC|\;\;bisects\;\;\angle BCD ...Yes\;\;because \\[3ex] \angle ACB = \angle CAB\;\;and \\[3ex] \angle CAB = \angle ACD \\[3ex] \therefore \angle ACB = \angle ACD ...equality\;\;with\;\;\angle CAB \\[3ex] |AC|\;\;bisects\;\;\angle BAD ...No \\[3ex] $Because there is no sufficient information to prove that$\angle CAD = \angle ACD$or Because there is no sufficient information to$\angle CAD = \angle CAB$(288.) ACT Radius$\overline{OA}$of the circle shown below is perpendicular to$\overline{AP}$The circle intersects$\overline{OP}$at B. The length of$\overline{AP}$is 12 centimeters, and the measure of ∠APO is 20°. Which of the following values is closest to the length, in centimeters, of$\overline{BP}$? (Note:$\sin 20^\circ \approx 0.342$,$\cos 20^\circ \approx 0.940$, and$\tan 20^\circ \approx 0.364$)$ A.\;\; 2.1 \\[3ex] B.\;\; 4.4 \\[3ex] C.\;\; 6.9 \\[3ex] D.\;\; 7.6 \\[3ex] E.\;\; 8.4 \\[3ex]  SOHCAHTOA \\[3ex] \tan 20 = \dfrac{\overline{OA}}{\overline{AP}} \\[5ex] 0.364 = \dfrac{\overline{OA}}{12} \\[5ex] \overline{OA} = 12 * 0.364 \\[3ex] \overline{OA} = 4.368 \\[3ex] \overline{OB} = \overline{OA} = 4.368\;cm ...radius \\[3ex] \cos 20 = \dfrac{\overline{AP}}{\overline{OP}} \\[5ex] 0.940 = \dfrac{12}{\overline{OP}} \\[5ex] 0.94 * \overline{OP} = 12 \\[3ex] \overline{OP} = \dfrac{12}{0.94} \\[5ex] \overline{OP} = 12.76595745 \\[3ex] \overline{OP} = \overline{OB} + \overline{BP} ...diagram \\[3ex] \overline{OB} + \overline{BP} = \overline{OP} \\[3ex] 4.368 + \overline{BP} = 12.76595745 \\[3ex] \overline{BP} = 12.76595745 - 4.368 \\[3ex] \overline{BP} = 8.397957447 \\[3ex] \overline{BP} \approx 8.4\;cm $(289.) GCSE A, B, C and D are points on the circumference of a circle, centre O. Angle BAD = 112° and angle DCO = 33°. (a) Show that angle y = 35° Give reasons for each stage of your working. (b) Work out angle z. Give reasons for your answer.$ (a) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle BCD + \angle BAD = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BCD + 112 = 180 \\[3ex] \angle BCD = 180 - 112 \\[3ex] \angle BCD = 68^\circ \\[3ex] \angle BCO + \angle DCO = \angle BCD ... diagram \\[3ex] \angle BCO + 33 = 68 \\[3ex] \angle BCO = 68 - 33 \\[3ex] \angle BCO = 35^\circ \\[3ex] y = \angle BCO = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle COB \\[3ex] (b) \\[3ex] \underline{\triangle COB} \\[3ex] \angle COB + \angle BCO + y = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle COB + 35 + 35 = 180 \\[3ex] \angle COB + 70 = 180 \\[3ex] \angle COB = 180 - 70 \\[3ex] \angle COB = 110^\circ \\[3ex] \angle COB = 2 * z ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 110 = 2z \\[3ex] 2z = 110 \\[3ex] z = \dfrac{110}{2} \\[5ex] z = 55^\circ $(290.) NYSED Regents Examination In the diagram below of circle O, chords$\overline{AB}$and$\overline{CD}$intersect at E If m$\overset{\huge\frown}{AC}$= 72° and m∠AEC = 58°, how many degrees are in m$\overset{\huge\frown}{DB}$? (1) 108° (3) 44° (2) 65° (4) 14°$ \angle AEC = \dfrac{\overset{\huge\frown}{AC} + \overset{\huge\frown}{DB}}{2} \\[5ex] ... Angle\;\;of\;\;Intersecting\;\;Chords\;\;Theorem \\[3ex] 58 = \dfrac{72 + \overset{\huge\frown}{DB}}{2} \\[5ex] 58(2) = 72 + \overset{\huge\frown}{DB} \\[3ex] 116 = 72 + \overset{\huge\frown}{DB} \\[3ex] 72 + \overset{\huge\frown}{DB} = 116 \\[3ex] \overset{\huge\frown}{DB} = 116 - 72 \\[3ex] \overset{\huge\frown}{DB} = 44^\circ $(291.) KCSE In the figure below, O is the centre of the circle. A, B, C and D are points on the circumference of the circle. Line AB is parallel to line DC and angle ADC = 55° Determine the size of angle ACB.$ \angle DAC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \angle DCA + \angle DAC + \angle ADC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DAC \\[3ex] \angle DCA + 90 + 55 = 180 \\[3ex] \angle DCA + 145 = 180 \\[3ex] \angle DCA = 180 - 145 \\[3ex] \angle DCA = 35^\circ \\[3ex] \angle BAC = \angle DCA = 35^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|DC| \;\;and\;\; |AB|\;\;are\;\;equal \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle D + \angle B = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle D = \angle ADC = 55^\circ ... diagram \\[3ex] \angle B = \angle ABC ...diagram \\[3ex] \implies \\[3ex] 55 + \angle ABC = 180 \\[3ex] \angle ABC = 180 - 55 \\[3ex] \angle ABC = 125^\circ \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ACB + \angle ABC + \angle BAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle ACB + 125 + 35 = 180 \\[3ex] \angle ACB + 160 = 180 \\[3ex] \angle ACB = 180 - 160 \\[3ex] \angle ACB = 20^\circ $(292.) ICSE AB and CD are two parallel chords of a circle such that AB = 24 cm and CD = 10 cm. If the radius of the circle is 13 cm, find the distance between the two chords. Construction: Draw the radii from centre O to these points on the circumference: point A point B point C point D$ radius = r = |OA| = |OB| = |OC| = |OD| = 13\;cm \\[3ex] ...given\;\;and\;\;diagram \\[3ex] |AN| = |NB| = \dfrac{|AB|}{2} = \dfrac{24}{2} = 12\;cm \\[3ex] ...line:\;ON\;\;\perp\;\;to\;\;chord:\;AB;\;\;bisects\;\;the\;\;chord \\[3ex] \underline{\triangle ONB} \\[3ex] |ON|^2 + |NB|^2 = |OB|^2 ... Pythagorean\;\;theorem \\[5ex] |ON|^2 + 12^2 = 13^2 \\[3ex] |ON|^2 + 144 = 169 \\[3ex] |ON|^2 = 169 - 144 \\[3ex] |ON|^2 = 25 \\[3ex] |ON| = \sqrt{25} \\[3ex] |ON| = 5 \\[5ex] |CM| = |MD| = \dfrac{|CD|}{2} = \dfrac{10}{2} = 5\;cm \\[3ex] ...line:\;OM\;\;\perp\;\;to\;\;chord:\;CD;\;\;bisects\;\;the\;\;chord \\[3ex] \underline{\triangle OMD} \\[3ex] |OM|^2 + |MD|^2 = |OD|^2 ... Pythagorean\;\;theorem \\[5ex] |OM|^2 + 5^2 = 13^2 \\[3ex] |OM|^2 + 25 = 169 \\[3ex] |OM|^2 = 169 - 25 \\[3ex] |OM|^2 = 144 \\[3ex] |OM| = \sqrt{144} \\[3ex] |OM| = 12 \\[5ex] \underline{Distance\;\;between\;\;chord\;|AB|\;\;and\;\;chord\;|CD|} \\[3ex] distance = |ON| + |OM| \\[3ex] distance = 5 + 12 \\[3ex] distance = 17\;cm $(293.) GCSE The points E, F and G lie on the circumference of a circle, centre O. JGH is a tangent to the circle. Find the value of x and the value of y.$ x + 58 = 90^\circ ...radius\;OG \perp tangent\;JGH\;\;at\;\;point\;\;of\;\;contact\;G \\[3ex] x = 90 - 58 \\[3ex] x = 32^\circ \\[5ex] y = 58^\circ ... \angle\;\;in\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;JGH\;\;and\;\;chord\;GF $(294.) NYSED Regents Examination In the diagram below, secants$\overline{RST}$and$\overline{RQP}$, drawn from point R, intersect circle O at S, T, Q, and P If RS = 6, ST = 4, and RP = 15, what is the length of$\overline{RQ}$?$ RS * RT = RQ * RP ...Intersecting\;\;Secants\;\;Theorem \\[3ex] 6 * (6 + 4) = RQ * 15 \\[3ex] 6 * 10 = RQ * 15 \\[3ex] RQ * 15 = 60 \\[3ex] RQ = \dfrac{60}{15} \\[5ex] |RQ| = 4 $(295.) NYSED Regents Examination Diameter$\overline{ROQ}$of circle O is extended through Q to point P, and tangent$\overline{PA}$is drawn. If m$\overset{\huge\frown}{RA}$= 100°, what is m∠P? (1) 10° (3) 40° (2) 20° (4) 50° Let us represent the information using a diagram. Be reminded that the measure of an arc is the angle subtended at the center by the arc.$ \angle ORA = \angle OAR = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ROA \\[3ex] \angle ORA + \angle OAR + \angle ROA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROA \\[3ex] p + p + 100 = 180 \\[3ex] 2p = 180 - 100 \\[3ex] 2p = 80 \\[3ex] p = \dfrac{80}{2} \\[5ex] p = 40 \\[3ex] \therefore \angle ORA = \angle OAR = 40^\circ \\[3ex] \angle OAP = 90^\circ ...radius\;OA \perp tangent\;PA\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle RAP = \angle OAR + \angle OAP ...diagram \\[3ex] \angle RAP = 40 + 90 = 130^\circ \\[3ex] \angle PRA = \angle ORA = 40^\circ ...diagram \\[3ex] \angle P = \angle RPA ...diagram \\[3ex] \angle RPA + \angle RAP + \angle PRA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RPA \\[3ex] \angle P + 130 + 40 = 180 \\[3ex] \angle P + 170 = 180 \\[3ex] \angle P = 180 - 170 \\[3ex] \angle P = 10^\circ $(296.) KCSE In the figure below, BOD is the diameter of the circle centre O. Angle ABD = 30° and angle AXD = 70° Determine the size of: (a) reflex angle BOC (b) angle ACO$ \underline{\triangle BAX} \\[3ex] 30 + \angle BAX = 70 \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle BAX = 70 - 30 \\[3ex] \angle BAX = 40^\circ \\[3ex] \angle BAC = \angle BAX = 40^\circ ... diagram \\[3ex] \angle BOC = 2 * \angle BAC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * 40 \\[3ex] \angle BOC = 80^\circ \\[3ex] (a) \\[3ex] Acute\;\angle BOC + Reflex\;\angle BOC = 360^\circ ... \angle s\;\;at\;\;a\;\;point \\[3ex] 80 + Reflex\;\angle BOC = 360 \\[3ex] Reflex\;\angle BOC = 360 - 80 \\[3ex] Reflex\;\angle BOC = 280^\circ \\[3ex] (b) \\[3ex] \angle BOC + \angle COX = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 80 + \angle COX = 180 \\[3ex] \angle COX = 180 - 80 \\[3ex] \angle COX = 100^\circ \\[3ex] \angle CXO = \angle AXD = 70^\circ ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle XCO} \\[3ex] \angle XCO + \angle COX + \angle CXO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle XCO + 100 + 70 = 180 \\[3ex] \angle XCO + 170 = 180 \\[3ex] \angle XCO = 180 - 170 \\[3ex] \angle XCO = 10 \\[3ex] \angle ACO = \angle XCO = 10^\circ ... diagram $(297.) NSC (297.1) Complete the following theorem: The exterior angle of a cyclic quadrilateral is equal to ... (297.2) The diagram below shows a circle PRLNM with centre O. ROM is a diameter and NL is produced to K. LN || RM$\hat{P_3} = 11^\circ$(297.2.1) Give a reason why$M\hat{P}R = 90^\circ$(297.2.2) Determine, giving reasons, THREE other angles each equal to 11° (297.2.3) Give a reason why LR = NM (297.2.4) Determine the size of$\hat{M_1}$(297.2.5) Determine the size of$\hat{L_1}$(297.1) The exterior angle of a cyclic quadrilateral is equal to the interior opppsite angle.$ (297.2.1) \\[3ex] M\hat{P}R = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] (297.2.2) \\[3ex] \hat{P_3} = 11^\circ ...given \\[3ex] \hat{M_2} = \hat{P_3} = 11^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \hat{L_4} = \hat{M_2} = 11^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|LN| \;\;and\;\; |RM|\;\;are\;\;equal \\[3ex] \hat{P_1} = \hat{L_4} = 11^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] $(297.2.3) LN || RM ...given Because parallel lines are equidistant from each other: LR = NM because each point on the line LN is the same distance as the corresponding point on the other line RM Hence the point L on the line LN is equidistant from the point R on the line RM Also, the point N on the line LN is equidistant from the point M on the line RM$ (297.2.4) \\[3ex] \hat{P_3} + \hat{P_2} + \hat{P_1} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] 11 + \hat{P_2} + 11 = 90 \\[3ex] \hat{P_2} + 22 = 90 \\[3ex] \hat{P_2} = 90 - 22 \\[3ex] \hat{P_2} = 68^\circ \\[3ex] \hat{M_1} = \hat{P_2} = 68^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (297.2.5) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;LRPN} \\[3ex] \hat{L_1} = \hat{P_3} + \hat{P_2} ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] \hat{L_1} = 11 + 68 \\[3ex] \hat{L_1} = 79^\circ $(298.) NYSED Regents Examination In circle M below, diameter$\overline{AC}$, chords$\overline{AB}$and$\overline{BC}$, and radius$\overline{MB}$are drawn. Which statement is not true? (1)$\triangle ABC$is a right triangle. (3) m$\overset{\huge\frown}{BC}$= m$\angle BMC$(2)$\triangle ABM$is isosceles. (4) m$\overset{\huge\frown}{AB}$=$\dfrac{1}{2}$m$\angle ACB$Let us analyze each option$ (1) \\[3ex] \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \rightarrow \triangle ABC \;\;is\;\;a\;\;right\;\;\triangle \\[3ex] YES \\[3ex] (2) \\[3ex] \angle MAB = \angle MBA ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ABM \\[3ex] YES \\[3ex] (3) \\[3ex] \overset{\huge\frown}{BC} = \angle BMC...intercepted\;\;\overset{\huge\frown}{} = central\;\;\angle \\[3ex] YES \\[3ex] (4) \\[3ex] \overset{\huge\frown}{AB} \ne \angle ACB \\[3ex] NO $(299.) KCSE In the figure below, P, Q, R and S are points on the circle with centre O. PRT and USTV are straight lines. Line USTV is a tangent to the circle at S. ∠RST = 50° and ∠RTV = 150° (a) Calculate the size of: (i) ∠ORS (ii) ∠USP (iii) ∠PQR (b) Given that RT = 7cm and ST = 9cm, calculate to 3 significant figures; (i) the length of line PR (ii) the radius of the circle.$ (a) \\[3ex] \angle SPR = \angle RST = 50^\circ ... \angle\;\;in\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;UST\;\;and\;\;chord\;SR \\[3ex] \angle SOR = 2 * \angle SPR ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle SOR = 2 * 50 \\[3ex] \angle SOR = 100^\circ \\[3ex] (i) \\[3ex] \angle ORS = \angle OSR = k ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle SOR \\[3ex] \angle ORS + \angle OSR + \angle SOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOR \\[3ex] k + k + 100 = 180 \\[3ex] 2k = 180 - 100 \\[3ex] 2k = 80 \\[3ex] k = \dfrac{80}{2} \\[5ex] k = 40 \\[3ex] \therefore \angle ORS = \angle OSR = 40^\circ \\[3ex] (ii) \\[3ex] \angle RTS + \angle RTV = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RTS + 150 = 180 \\[3ex] \angle RTS = 180 - 150 \\[3ex] \angle RTS = 30^\circ \\[3ex] \angle PRS = \angle RTS + \angle RST ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle PRS = 30 + 50 \\[3ex] \angle PRS = 80^\circ \\[3ex] \angle USP = \angle PRS = 80^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle SPR} \\[3ex] \angle PSR + \angle PRS + \angle SPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle PSR + 80 + 50 = 180 \\[3ex] \angle PSR + 130 = 180 \\[3ex] \angle PSR = 180 - 130 \\[3ex] \angle PSR = 50^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\; PQRS} \\[3ex] \angle Q + \angle S = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle PQR = \angle Q ... diagram \\[3ex] \angle S = \angle PSR = 50^\circ ... diagram \\[3ex] \implies \\[3ex] \angle PQR + 50 = 180 \\[3ex] \angle PQR = 180 - 50 \\[3ex] \angle PQR = 130^\circ \\[3ex] (b) \\[3ex] $We can solve (b) using at least two approaches. Use any approach you prefer. For KCSE students, use the 1st approach.$ (i) \\[3ex] \underline{1st\;\;Approach:\;\;Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] |RT| * |PT| = |ST|^2 ...Intersecting\;\;Secant-Tangent\;\;Theorem \\[3ex] |PT| = |PR| + |RT| ... diagram \\[3ex] \implies \\[3ex] |RT|(|PR| + |RT|) = |ST|^2 \\[3ex] 7(|PR| + 7) = 9^2 \\[3ex] 7(|PR| + 7) = 81 \\[3ex] |PR| + 7 = \dfrac{81}{7} \\[5ex] |PR| + 7 = 11.57142857 \\[3ex] |PR| = 11.57142857 - 7 \\[3ex] |PR| = 4.571428571 \\[3ex] |PR| \approx 4.57\;cm...to\;\;3\;\;significant\;\;figures \\[3ex] (ii) \\[3ex] \underline{2nd\;\;Approach:\;\;Cosine\;\;Law\;\;and\;\;Sine\;\;Law} \\[3ex] |SR|^2 = |RT|^2 + |ST|^2 - 2 * |RT| * |ST| * \cos \angle RTS ...Cosine\;\;Law \\[3ex] |SR|^2 = 7^2 + 9^2 - 2(7)(9) \cos 30 \\[3ex] = 49 + 81 - 126(0.8660254038) \\[3ex] = 130 - 109.1192009 \\[3ex] = 20.88079912 \\[3ex] |SR| = \sqrt{20.88079912} \\[3ex] |SR| = 4.569551304 \\[3ex] \underline{\triangle PSR} \\[3ex] \dfrac{||PR|}{\sin \angle PSR} = \dfrac{|SR|}{\sin \angle SPR}...Sine\;\;Law \\[5ex] \dfrac{|PR|}{\sin 50} = \dfrac{4.569551304}{\sin 50} \\[5ex] same\;\;denominator;\;\;equate\;\;the\;\;numerators \\[3ex] |PR| = 4.569551304 \\[3ex] |PR| \approx 4.57\;cm...to\;\;3\;\;significant\;\;figures \\[3ex] (ii) \\[3ex] |OR| = |OS| = radius = r \\[3ex] \underline{First\;\;Approach:\;\;Sine\;\;Law} \\[3ex] \dfrac{||OR|}{\sin \angle OSR} = \dfrac{|SR|}{\sin \angle SOR}...Sine\;\;Law \\[5ex] \dfrac{r}{\sin 40} = \dfrac{4.569551304}{\sin 100} \\[5ex] r = \dfrac{4.569551304 * \sin 40}{\sin 100} \\[5ex] r = \dfrac{2.93725096}{0.984807753} \\[5ex] r = 2.982562791 \\[3ex] r \approx 2.98\;cm ...to\;\;3\;\;significant\;\;figures \\[3ex] \underline{Second\;\;Approach:\;\;Cosine\;\;Law} \\[3ex] |SR|^2 = |OR|^2 + |OS|^2 - 2 * |OR| * |OS| * \cos \angle SOR ...Cosine\;\;Law \\[3ex] 20.88079912 = r^2 + r^2 - 2 * r * r * \cos 100 \\[3ex] 20.88079912 = 2r^2 - 2r^2 * -0.1736481777 \\[3ex] 20.88079912 = 2r^2 + 0.3472963553r^2 \\[3ex] 20.88079912 = 2.347296355r^2 \\[3ex] 3.347296355r^2 = 20.88079912 \\[3ex] r^2 = \dfrac{20.88079912}{2.347296355} \\[5ex] r^2 = 8.8956808 \\[3ex] r = \sqrt{8.8956808} \\[3ex] r = 2.982562791 \\[3ex] r \approx 2.98\;cm ...to\;\;3\;\;significant\;\;figures $(300.) GCSE In this question, all lengths are in centimetres. A is a point on a circle, centre O. B is a point on a different circle, centre O. AB = 20 The equation of the larger circle is$x^2 + y^2 = 144$radius of smaller circle : radius of larger circle = 4 : 5$ \underline{larger\;\;circle} \\[3ex] radius = |OB| \\[3ex] Equation:\;\; x^2 + y^2 = 144 \\[3ex] Compare:\;\;x^2 + y^2 = r^2 \\[3ex] r^2 = 144 \\[3ex] r = +\sqrt{144} \\[3ex] r = 12 \\[3ex] |OB| = 12 \\[3ex] \underline{smaller\;\;circle} \\[3ex] radius = |OA| \\[3ex] \underline{Ratio\;\;of\;\;Radii} \\[3ex] |OA| : |OB| = 4 : 5 \\[3ex] |OA| : 12 = 4 : 5 \\[3ex] \dfrac{|OA|}{12} = \dfrac{4}{5} \\[5ex] |OA| = \dfrac{12(4)}{5} \\[5ex] |OA| = 9.6 \\[3ex] \underline{\triangle AOB} \\[3ex] |AB|^2 = |OA|^2 + |OB|^2 - 2 * |OA| * |OB| \cos \angle AOB \\[3ex] ... Cosine\;\;Law \\[3ex] 20^2 = 9.6^2 + 12^2 - 2(9.6)(12) \cos \angle AOB \\[3ex] 400 = 92.16 + 144 - 230.4 * \cos \angle AOB \\[3ex] 400 = 236.16 - 230.4 * \cos \angle AOB \\[3ex] 230.4 * \cos \angle AOB = 236.16 - 400 \\[3ex] 230.4 * \cos \angle AOB = -163.84 \\[3ex] \cos \angle AOB = -\dfrac{163.84}{230.4} \\[5ex] \cos \angle AOB = -0.7111111111 \\[3ex] \angle AOB = \cos^{-1}{-0.7111111111} \\[3ex] \angle AOB = 135.3253904^\circ $(301.) NSC In the diagram, TSR is a diameter of the larger circle having centre S. Chord TQ of the larger circle cuts the smaller circle at M. PQ is a common tangent to the two circles at Q. SQ is drawn. RP ⊥ PQ and MS || QR. (301.1) Prove, giving reasons that: (301.1.1) SQ is the diameter of the smaller circle. (301.1.2)$RT = \dfrac{RQ^2}{RP}$(301.2) If MS = 14 units and PQ =$\sqrt{640}$units, calculate, giving reasons, the length of the radius of the larger circle.$ (301.1.1) \\[3ex] TRS\;\;is\;\;a\;\;diameter \\[3ex] \angle TQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \hat{M_1} = \angle TQR = 90^\circ ...corresponding\;\;\angle s\;\;between\;\;parallel\;\;lines:|MS| \;\;and\;\; |QR|\;\;are\;\;equal \\[3ex] \hat{M_1} + \hat{M_2} = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + \hat{M_2} = 180 \\[3ex] \hat{M_2} = 180 - 90 \\[3ex] \hat{M_2} = 90^\circ \\[3ex] \implies \\[3ex] \hat{M_2} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \therefore SQ \;\;is\;\;a\;\;diameter \\[3ex] (301.1.2) \\[3ex] \hat{Q_3} = \hat{T} ...\angle\;\;between\;\;tangent\;QP\;\;and\;\;chord\;QR = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \implies \\[3ex] \sin \hat{Q_3} = \sin \hat{T} \\[3ex] \underline{\triangle QRP} \\[3ex] \sin \hat{Q_3} = \dfrac{RP}{RQ} ...SOHCAHTOA \\[5ex] \underline{\triangle QTR} \\[3ex] \sin \hat{T} = \dfrac{RQ}{RT} ...SOHCAHTOA \\[5ex] \implies \\[3ex] \dfrac{RP}{RQ} = \dfrac{RQ}{RT} \\[5ex] |RP| * |RT| = |RQ|^2 \\[3ex] |RT| = \dfrac{|RQ|^2}{|RP|} \\[5ex] $To solve question (301.2), knowledge of Triangle Theorems is required.$ radius\;\;of\;\;the\;\;larger\;\;circle = r = |RS| = |ST| = \dfrac{|RT|}{2} ...diagram \\[5ex] \underline{\triangle QTR} \\[3ex] |MS| = \dfrac{|QR|}{2} ...Converse\;\;of\;\;the\;\;Midpoint\;\;Theorem \\[5ex] 14 = \dfrac{|QR|}{2} \\[5ex] |QR| = 14(2) \\[3ex] |RQ| = |QR| = 28\;units \\[3ex] \underline{\triangle QRP} \\[3ex] |QR|^2 = |PQ|^2 + |RP|^2 ...Pythagorean\;\;Theorem \\[3ex] 28^2 = (\sqrt{640})^2 + |RP|^2 \\[3ex] 784 = 640 + |RP|^2 \\[3ex] 640 + |RP|^2 = 784 \\[3ex] |RP|^2 = 784 - 640 \\[3ex] |RP|^2 = 144 \\[3ex] |RP| = \sqrt{144} \\[3ex] |RP| = 12\;units \\[3ex] |RT| = \dfrac{|RQ|^2}{|RP|} \\[5ex] |RT| = \dfrac{28^2}{12} \\[5ex] |RT| = \dfrac{784}{12} \\[5ex] r = \dfrac{|RT|}{2} = \dfrac{784}{12} \div 2 \\[5ex] r = \dfrac{784}{12} * \dfrac{1}{2} \\[5ex] r = \dfrac{392}{12} \\[5ex] r = \dfrac{98}{3}\;units $(302.) NYSED Regents Examination Quadrilateral ABCD is inscribed in circle O, as shown below. If m∠A = 80°, and m∠B = 75°, m∠C = (y + 30)°, and m∠D = (x - 10)°, which statement is true? (1) x = 85 and y = 50 (3) x = 110 and y = 75 (2) x = 90 and y = 45 (4) x = 115 and y = 70$ (x - 10) + 75 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] x - 10 + 75 = 180 \\[3ex] x + 65 = 180 \\[3ex] x = 180 - 65 \\[3ex] x = 115^\circ \\[3ex] Similarly \\[3ex] (y + 30) + 80 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] y + 30 + 80 = 180 \\[3ex] y + 110 = 180 \\[3ex] y = 180 - 110 \\[3ex] y = 70^\circ $(303.) ACT Points A, B, D, and E lie on the circle shown below. Secants$\overleftrightarrow{AC}$and$\overleftrightarrow{CE}$intersect at C. The chords$overline{AB}$and$\overline{DE}$are congruent. Minor arc$\overset{\huge\frown}{AE}$measures 106° Minor arc$\overset{\huge\frown}{BD}$measures 64° What is the measure of minor arc$\overset{\huge\frown}{DE}$?$ F.\;\; 42^\circ \\[3ex] G.\;\; 84^\circ \\[3ex] H.\;\; 85^\circ \\[3ex] J.\;\; 90^\circ \\[3ex] K.\;\; 95^\circ \\[3ex] $Let us represent the information in the circle Construction: (a.) Let O be the center of the circle. (b.) Indicate the central angles corresponding to the minor arcs. (c.) Indicate the congruent lines (chords)$ \underline{\triangle AOE} \\[3ex] \angle OAE = \angle OEA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOE \\[3ex] \angle OAE + \angle OEA + \angle AOE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOE \\[3ex] p + p + 106 = 180 \\[3ex] 2p = 180 - 106 \\[3ex] 2p = 74 \\[3ex] p = \dfrac{74}{2} \\[5ex] p = 37 \\[3ex] \therefore \angle OAE = \angle OEA = 37^\circ \\[3ex] \angle ACE = \dfrac{\overset{\huge\frown}{AE} - \overset{\huge\frown}{BD}}{2} ... Angle\;\;of\;\;Intersecting\;\;Chords\;\;Theorem \\[5ex] \implies \\[3ex] \angle ACE = \dfrac{\angle AOE - \angle BOD}{2} \\[5ex] \angle ACE = \dfrac{106 - 64}{2} \\[5ex] \angle ACE = \dfrac{42}{2} \\[5ex] \angle ACE = 21^\circ \\[3ex] \underline{\triangle ACE} \\[3ex] |AB| \cong |DE| ...given \\[3ex] \implies \\[3ex] \angle CAE = \angle CEA = k \\[3ex] \angle CAE + \angle CEA + \angle ACE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACE \\[3ex] k + k + 21 = 180 \\[3ex] 2k = 180 - 21 \\[3ex] 2k = 159 \\[3ex] k = \dfrac{159}{2} \\[5ex] k = 79.5 \\[3ex] \therefore \angle CAE = \angle CEA = 79.5^\circ \\[3ex] \angle OEA + \angle OED = \angle CEA ...diagram \\[3ex] 37 + \angle OED = 79.5 \\[3ex] \angle OED = 79.5 - 37 \\[3ex] \angle OED = 42.5^\circ \\[3ex] \underline{\triangle EOD} \\[3ex] \angle OED = \angle ODE = 42.5^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EOD \\[3ex] \angle OED + \angle ODE + \angle EOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOD \\[3ex] 42.5 + 42.5 + \angle EOD = 180 \\[3ex] 85 + \angle EOD = 180 \\[3ex] \angle EOD = 180 - 85 \\[3ex] \angle EOD = 95^\circ \\[3ex] \implies \\[3ex] \overset{\huge\frown}{DE} = \angle EOD = 85^\circ $(304.) KCSE In the figure below, OS is the radius of the circle centre O. Chords SQ and TU are extended to meet at P and OR is perpendicular to QS at R. OS = 61cm, PU = 50cm, UT = 40cm and PQ = 30cm. (a) Calculate the length of: (i) QS; (ii) OR. (b) Calculate, correct to 1 decimal place: (i) the size of angle ROS; (ii) the length of the minor arc QS.$ (a) \\[3ex] (i) \\[3ex] \underline{Intersecting\;\;Secants\;\;Theorem} \\[3ex] PU * PT = PQ * PS \\[3ex] PT = PU + UT \\[3ex] PT = 50 + 40 \\[3ex] PT = 90\;cm \\[3ex] PS = PQ + QS ... diagram \\[3ex] \implies \\[3ex] PU * PT = PQ(PQ + QS) \\[3ex] 50 * 90 = 30(30 + QS) \\[3ex] 30(30 + QS) = 50 * 90 \\[3ex] 30 + QS = \dfrac{50 * 90}{30} \\[5ex] 30 + QS = 150 \\[3ex] QS = 150 - 30 \\[3ex] QS = 120\;cm \\[3ex] (ii) \\[3ex] QR = RS = \dfrac{QS}{2} = \dfrac{120}{2} = 60\;cm \\[5ex] ...line\;OR\;\;\perp\;\;to\;\;chord\;QS;\;\;bisects\;\;the\;\;chord \\[3ex] \underline{\triangle ORS} \\[3ex] |OR|^2 + |RS|^2 = |OS|^2 ...Pythagorean\;\;Theorem \\[3ex] |OR|^2 + 60^2 = 61^2 \\[3ex] |OR|^2 = 61^2 - 60^2 \\[3ex] = (61 + 60)(61 - 60) \\[3ex] = 121(1) \\[3ex] = 121 \\[3ex] |OR| = \sqrt{121} \\[3ex] |OR| = 11\;cm \\[3ex] (b) \\[3ex] let\;\;\angle ROS = \alpha \\[3ex] SOHCAHTOA \\[3ex] (i) \\[3ex] \sin \alpha = \dfrac{RS}{OS} \\[5ex] \sin \alpha = \dfrac{60}{61} \\[5ex] \sin \alpha = 0.9836065574 \\[3ex] \alpha = \sin^{-1}(0.9836065574) \\[3ex] \alpha = 79.61114218 \\[3ex] \therefore \angle ROS \approx 79.6^\circ ...to\;\;1\;\;decimal\;\;place \\[3ex] (ii) \\[3ex] QO = OS = r = 61\;cm...radii \\[3ex] \angle QOR = \angle ROS = \alpha \\[3ex] ...line\;OR\;\;\perp\;\;to\;\;chord\;QS;\;\;bisects\;\;its\;\;arc \\[3ex] let\;\;\angle QOS = \beta \\[3ex] \beta = 2 * \alpha \\[3ex] \beta = 2(79.61114218) \\[3ex] \beta = 159.2222844^\circ \\[3ex] \overset{\huge\frown}{QS} = \dfrac{\beta}{360} * 2\pi r \\[5ex] \overset{\huge\frown}{QS} = \dfrac{159.2222844}{360} * 2 * \dfrac{22}{7} * 61 \\[5ex] \overset{\huge\frown}{QS} = \dfrac{427352.6112}{2520} \\[5ex] \overset{\huge\frown}{QS} = 169.5843695 \\[3ex] \overset{\huge\frown}{QS} \approx 169.6\;cm $(305.) NZQA The points P, R, and S all lie on the circumference of a circle, with centre C. The line CP is parallel to the line RS. Angle CPR = 25° (i) Find the size, x, of angle RCP. Justify your answer. (ii) Find the size, y, of angle RCS. Justify your answer.$ (i) \\[3ex] \underline{\triangle RCP} \\[3ex] \angle CRP = \angle CPR = 25^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle RCP \\[3ex] \angle RCP + \angle CRP + \angle CPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RCP \\[3ex] x + 25 + 25 = 180 \\[3ex] x + 50 = 180 \\[3ex] x = 180 - 50 \\[3ex] x = 30^\circ \\[5ex] (ii) \\[3ex] \angle PRS = 25^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|CP| \;\;and\;\; |RS|\;\;are\;\;equal \\[3ex] \angle CRS = \angle CRP + \angle PRS ...diagram \\[3ex] \angle CRS = 25 + 25 \\[3ex] \angle CRS = 50^\circ \\[3ex] \underline{\triangle RCS} \\[3ex] \angle CSR = \angle CRS = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle RCS \\[3ex] \angle RCS + \angle CSR + \angle CRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RCS \\[3ex] y + 50 + 50 = 180 \\[3ex] y + 100 = 180 \\[3ex] y = 180 - 100 \\[3ex] y = 80^\circ \\[5ex] $(306.) GCSE In this question, all lengths are in centimetres. The diagram shows a sketch of a circle, centre O. Points A, B, C, D, E and F lie on the circumference of the circle. Triangles AOB, BOC, COD, DOE, EOF and FOA are congruent. The circle has equation$x^2 + y^2 = \dfrac{25}{4}$Calculate the perimeter of the hexagon ABCDEF. You must justify any decisions that you make.$ \underline{Circle} \\[3ex] Equation:\;\; x^2 + y^2 = \dfrac{25}{4} \\[5ex] Compare:\;\; x^2 + y^2 = r^2 \\[3ex] \implies \\[3ex] r^2 = \dfrac{25}{4} \\[5ex] r = +\sqrt{\dfrac{25}{4}} \\[5ex] r = \dfrac{5}{2}\;cm \\[5ex] |OA| = |OB| = |OC| = |OD| = |OE| = |OF| = radius = r = \dfrac{5}{2}\;cm \\[5ex] \underline{Hexagon} \\[3ex] Sum\;\;of\;\;interior\;\;\angle s \\[3ex] = 180(n - 2) \\[3ex] = 180(6 - 2)...6-sided\;\;polygon \\[3ex] = 180(4) \\[3ex] Each\;\;interior\;\;\angle = \dfrac{Sum}{n} \\[5ex] = \dfrac{180(4)}{6} \\[5ex] = 30(4) \\[3ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle FAB = \angle ABC = \angle BCD = \angle CDE = \angle DEF = \angle EFA = 120^\circ \\[3ex] \angle OFA = \angle OAF = \dfrac{120}{2} = 60^\circ...congruent\;\;\triangle s \\[5ex] The\;\;same\;\;applies\;\;to\;\;all\;\;other\;\;congruent\;\;\triangle s \\[3ex] \implies \\[3ex] each\;\;\triangle \;\;is\;\;an\;\;equilateral\;\;\triangle...each\;\;\angle\;\;of\;\;60^\circ \\[3ex] \implies \\[3ex] \underline{Sides\;\;of\;\;the\;\;Hexagon} \\[3ex] |AB| = |BC| = |CD| = |DE| = |EF| = |FA| = \dfrac{5}{2}\;cm \\[5ex] Perimeter\;\;of\;\;Hexagon = 6\left(\dfrac{5}{2}\right) ... 6\;\;sides \\[5ex] Perimeter = 3(5) \\[3ex] Perimeter = 15\;cm $(307.) CSEC (a.) W, X and U are points on the circumference of a circle. TV is a tangent to the circle at U. UW is a diameter of the circle and triangle WXU is isosceles. Using appropriate theorems, state THREE reasons that explain why the measure of Angle z is 45°. (b.) The diagram below shows a circle with diameter KF. Line EFG is a tangent to the circle at F. The points F, H, K and J lie on the circumference of the circle. By showing EACH step in your work, where appropriate, find the value for EACH of the following angles: (i.) Angle x (ii.) Angle y$ (a.) \\[3ex] \angle WXU = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle XWU = \angle XUW = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle WXU \\[3ex] \angle XWU + \angle XUW + \angle WXU = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WXU \\[3ex] \implies \\[3ex] p + p + 90 = 180 \\[3ex] 2p = 180 - 90 \\[3ex] 2p = 90 \\[3ex] p = \dfrac{90}{2} \\[5ex] p = 45 \\[3ex] \therefore \angle XWU = \angle XUW = 45^\circ \\[3ex] \underline{First\;\;Reason} \\[3ex] z = \angle XWU = 45^\circ ...\angle\;\;between\;\;tangent\;TV\;\;and\;\;chord\;XU = \angle\;\;in\;\;alternate\;\;segment \\[5ex] \underline{Second\;\;Reason} \\[3ex] UW = diameter...given \\[3ex] Let\;\;C = center\;\;of\;\;the\;\;circle \\[3ex] \implies \\[3ex] WC = CU = radius \\[3ex] \angle WUV = 90^\circ ...radius\;CU \perp tangent\;TUV\;\;at\;\;point\;\;of\;\;contact\;U \\[3ex] \angle WUV = \angle XUW + z...diagram \\[3ex] \implies \\[3ex] \angle XUW + z = 90 \\[3ex] 45 + z = 90 \\[3ex] z = 90 - 45 \\[3ex] z = 45^\circ \\[5ex] \underline{Third\;\;Reason} \\[3ex] \angle WUT = 90^\circ ...radius\;CU \perp tangent\;TUV\;\;at\;\;point\;\;of\;\;contact\;U \\[3ex] \angle WUT + \angle XUW + z = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + 45 + z = 180 \\[3ex] 135 + z = 180 \\[3ex] z = 180 - 135 \\[3ex] z = 45^\circ \\[5ex] (b.) \\[3ex] (i.) \\[3ex] x = 56^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii.) \\[3ex] y = x + 19 ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] y = 56 + 19 \\[3ex] y = 75^\circ $(308.) KCSE In the figure below, O is the centre of the circle. PQRS is a cyclic quadrilateral and RST is a straight line. Angle RPQ = 21° and angle TSP = 147° Calculate the size of angle SRQ.$ \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \angle RQP = \angle TSP = 147^\circ \\[3ex] ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quad = interior\;\;opposite\;\;\angle \\[3ex] \underline{\triangle RQP} \\[3ex] \angle PRQ + \angle RQP + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle PRQ + 147 + 21 = 180 \\[3ex] \angle PRQ + 168 = 180 \\[3ex] \angle PRQ = 180 - 168 \\[3ex] \angle PRQ = 12^\circ \\[3ex] \angle SRP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \angle SRQ = \angle SRP + \angle PRQ ... diagram \\[3ex] \angle SRQ = 90 + 12 \\[3ex] \angle SRQ = 102^\circ $(309.) curriculum.gov.mt (a.) WXYZ is a cyclic quadrilateral. XY and WZ are parallel. ∠WXY = 84°. Work out the size of ∠XYZ. (b.) Points B, C, D and E lie on the circumference of circle centre O and line AE is a tangent to the circle at E. ∠BDE = 35° Work out the value of the angles marked x, y and z, giving reasons for your answers. (a.) For us to use the one of the properties of angles between parallel lines: Construction: Extend Line YX to A Extend Line ZW to B$ (a.) \\[3ex] \angle XWB = \angle WXY = 84^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle XYZ = \angle XWB = 84^\circ ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[5ex] (b.) \\[3ex] \angle ODE = \angle BDE = \angle ADE = 35^\circ ...diagram \\[3ex] \angle OED = \angle ODE = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EOD \\[3ex] \angle OEA = 90^\circ ...radius\;OE \perp tangent\;AE\;\;at\;\;point\;\;of\;\;contact\;E \\[3ex] \underline{\triangle ADE} \\[3ex] \angle AED = \angle OEA + \angle OED ...diagram \\[3ex] \angle AED = 90 + 35 = 125^\circ \\[3ex] z + \angle AED + \angle ADE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] z + 125 + 35 = 180 \\[3ex] z + 160 = 180 \\[3ex] z = 180 - 160 \\[3ex] z = 20^\circ \\[5ex] y = \angle ODE + \angle OED ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] y = 35 + 35 \\[3ex] y = 70^\circ \\[3ex] OR \\[3ex] \underline{\triangle AOE} \\[3ex] z + y + \angle OEA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] 20 + y + 90 = 180 \[3ex] y + 110 = 180 \\[3ex] y = 180 - 110 \\[3ex] y = 70^\circ \\[5ex] y = 2x ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2x = y \\[3ex] 2x = 70 \\[3ex] x = \dfrac{70}{2} \\[5ex] x = 35^\circ \\[3ex] OR \\[3ex] x = 35^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $(310.) ICSE In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons the measure of: (i) ∠BCD (ii) ∠BOD (iii) ∠OBD$ (i) \\[3ex] \angle BCD = \angle DAE = 70^\circ \\[3ex] ... exterior\;\; \angle \:\:of\:\;cyclic\:\:quad.\;ABCD = interior\;\;opposite\;\;\angle \\[3ex] (ii) \\[3ex] \angle BOD = 2 * \angle BCD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOD = 2 * 70 \\[3ex] \angle BOD = 140^\circ \\[3ex] (iii) \\[3ex] \angle OBD = \angle ODB = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OBD \\[3ex] \angle OBD + \angle ODB + \angle BOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OBD \\[3ex] \implies \\[3ex] p + p + 140 = 180 \\[3ex] 2p = 180 - 140 \\[3ex] 2p = 40 \\[3ex] p = \dfrac{40}{2} \\[5ex] p = 20 \\[3ex] \therefore \angle OBD = \angle ODB = 20^\circ $(311.) GCSE The diagram shows a circle, centre O, radius 8cm. Chord RT is 12cm and ATB is a tangent to the circle at T (a) Use trigonometry to calculate angle ROT. (b) Find angle RTB (c) Calculate the length of arc RT$ (a) \\[3ex] |OT| = |OR| = radius = r = 8\;cm \\[3ex] \underline{\triangle ROT} \\[3ex] |TR|^2 = |OT|^2 + |OR|^2 - 2 * |OT| * |OR| * \cos\angle ROT ...Cosine\;\;Law \\[3ex] 12^2 = 8^2 + 8^2 - 2(8)(8) \cos\angle ROT \\[3ex] 144 = 64 + 64 - 128\cos\angle ROT \\[3ex] 144 = 128 - 128\cos \angle ROT \\[3ex] 144 = 128(1 - \cos\angle ROT) \\[3ex] \dfrac{144}{128} = 1 - \cos\angle ROT \\[5ex] 1.125 = 1 - \cos \angle ROT \\[3ex] \cos \angle ROT = 1 - 1.125 \\[3ex] \cos \angle ROT = -0.125 \\[3ex] \angle ROT = \cos^{-1}(-0.125) \\[3ex] \angle ROT = 97.18075578^\circ \\[3ex] (b) \\[3ex] \angle OTR = \angle ORT = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ROT \\[3ex] \angle OTR + \angle ORT + \angle ROT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROT \\[3ex] p + p + 97.18075578 = 180 \\[3ex] 2p = 180 - 97.18075578 \\[3ex] 2p = 82.81924422 \\[3ex] p = \dfrac{82.81924422}{2} \\[5ex] p = 41.40962211 \\[3ex] \therefore \angle OTR = \angle ORT = 41.40962211^\circ \\[3ex] \angle OTB = 90^\circ ...radius\;OT \perp tangent\;ATB\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle OTR + \angle RTB = \angle OTB ... diagram \\[3ex] 41.40962211 + \angle RTB = 90 \\[3ex] \angle RTB = 90 - 41.40962211 \\[3ex] \angle RTB = 48.59037789^\circ \\[3ex] (c) \\[3ex] Length\;\;of\;\;arc\;RT = \dfrac{\angle ROT}{360} * 2\pi * r \\[5ex] = \dfrac{97.18075578}{360} * 2 * \dfrac{22}{7} * 8 \\[5ex] = \dfrac{34207.62603}{2520} \\[5ex] = 13.57445478\;cm $(312.) WASSCE Find the value of y in the diagram.$ A.\:\: 10 \\[3ex] B.\:\: 15 \\[3ex] C.\:\: 20 \\[3ex] D.\:\: 30 \\[3ex]  x = 2y ...\angle s \:\:in\:\:the\:\:same\:\:segment\;\;are\;\;equal \\[3ex] 60 = 2y + 2x \\[3ex] exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] Substitute\;\;2y\;\;for\;\;x \\[3ex] 60 = 2y + 2(2y) \\[3ex] 60 = 2y + 4y \\[3ex] 60 = 6y \\[3ex] 6y = 60 \\[3ex] y = \dfrac{60}{6} \\[5ex] y = 10^\circ $(313.) (314.) KCSE Two chords AB and CD of a circle intersect internally at T. AB = 15 cm, TB = 3 cm and TD = 4 cm. Find the ratio in which T divides CD. Construction: Draw the circle to represent the information.$ |AT| + |TB| = |AB| ...diagram \\[3ex] |AT| + 3 = 15 \\[3ex] |AT| = 15 - 3 \\[3ex] |AT| = 12\;cm \\[3ex] \underline{Intersecting\;\;Chords\;\;Theorem} \\[3ex] |CT| * |TD| = |AT| * |TB| \\[3ex] |CT| * 4 = 12 * 3 \\[3ex] |CT| = \dfrac{12 * 3}{4} \\[5ex] |CT| = 9\;cm \\[3ex] Ratio\;\;in\;\;which\;\;T\;\;divides\;\;CD \\[3ex] = |CT| : |TD| \\[3ex] = 9 : 4 $(315.) (316.) GCSE The diagram shows a circle with centre O. Points A, B, C and D all lie on the circumference of th circle. The radius of the circle is 3.6cm, BC = 4.1 cm and$B\hat{C}D$= 93°. Prove that$D\hat{B}C$= 52.3°, correct to 3 significant figures. You must show all your working and give a reason for each stage of your proof.$ radius = |AO| = |OC| = 3.6\;cm ... diagram \\[3ex] diameter = |AC| = 2 * radius = 2(3.6) = 7.2\;cm \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \dfrac{\sin \angle BAC}{|BC|} = \dfrac{\sin \angle ABC}{|AC|} ... Sine\;\;Law \\[5ex] \dfrac{\sin \angle BAC}{4.1} = \dfrac{\sin 90}{7.2} \\[5ex] \sin \angle BAC = \dfrac{4.1 * \sin 90}{7.2} \\[5ex] \sin \angle BAC = 0.5694444444 \\[3ex] \angle BAC = \sin^{-1}(0.5694444444) \\[3ex] \angle BAC = 34.71149426^\circ \\[3ex] \angle BDC = \angle BAC = 34.71149426^\circ \\[3ex] ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle BCD} \\[3ex] \angle DBC + \angle BCD + \angle BDC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle DBC + 93 + 34.71149426 = 180 \\[3ex] \angle DBC + 127.7114943 = 180 \\[3ex] \angle DBC = 180 - 127.7114943 \\[3ex] \angle DBC = 52.28850574 \\[3ex] \angle DBC \approx 52.3^\circ ...to\;\;3\;\;significant\;\;figures $(317.) (318.) (319.) NZQA The points P, Q, R, and S all lie on the circumference of a circle, with centre C. Angle PQR = 72° Angle PRS = 28° Find the size, y, of angle SPR. Justify your answer.$ \angle PSR + \angle PQR = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle PSR + 72 = 180 \\[3ex] \angle PSR = 180 - 72 \\[3ex] \angle PSR = 108^\circ \\[3ex] \angle SPR + \angle PSR + \angle PRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] y + 108 + 28 = 180 \\[3ex] y + 136 = 180 \\[3ex] y = 180 - 136 \\[3ex] y = 44^\circ $(320.) ACT Graphed in the standard (x, y) coordinate plane below is line l and the circle with equation$(x - 2)^2 + y^2 = 1$. Line l passes through O(0, 0) and is tangent to the circle at A, and B is the center of the circle. What is the measure of$\angle AOB$?$ F.\;\; 15^\circ \\[3ex] G.\;\; 22.5^\circ \\[3ex] H.\;\; 30^\circ \\[3ex] J.\;\; 45^\circ \\[3ex] K.\;\; 60^\circ \\[3ex]  \underline{Equation\;\;of\;a\;\;circle\;\;with\;\;center\;(h, k)} \\[3ex] (x - k)^2 + (y - k)^2 = r^2 \\[3ex] Compare\;\;with \\[3ex] (x - 2)^2 + y^2 = 1 \\[3ex] \implies \\[3ex] (x - 2)^2 + (y - 0)^2 = 1^2 \\[3ex] center = (h, k) = (2, 0) \\[3ex] radius = r = 1\;unit \\[3ex] r = |AB| = 1\;unit ...diagram \\[3ex] |OB| = distance\;\;between\;\;O(0, 0)\;\;and\;\;B(2, 0) \\[3ex] |OB| = \sqrt{(2 - 0)^2 + (0 - 0)^2}... Distance\;\;Formula \\[3ex] |OB| = \sqrt{2^2 + 0} \\[3ex] |OB| = \sqrt{4} \\[3ex] |OB| = 2\;units \\[3ex] \dfrac{\sin \angle AOB}{|AB|} = \dfrac{\sin \angle OAB}{|OB|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle AOB}{1} = \dfrac{\sin 90}{2} \\[5ex] \sin \angle AOB = \dfrac{1}{2} \\[5ex] \angle AOB = \sin^{-1}\left(\dfrac{1}{2}\right) \\[5ex] \angle AOB = 30^\circ $(321.) GCSE L, M and N are points on a circle, centre O. QMT is the tangent to the circle at M. (a) (i) Find the size of angle NOM. (ii) Give a reason for your answer. (b) (i) Find the size of angle NMQ. (ii) GIve a reason for your answer.$ (a) \\[3ex] \angle NOM = 2 * \angle NLM ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle NOM = 2 * 27 \\[3ex] \angle NOM = 54^\circ \\[3ex] (b) \\[3ex] $There are at least two approaches of doing (b) Use any approach you prefer.$ \underline{First\;\;Approach} \\[3ex] \angle NMQ = 27^\circ ...\angle\;\;between\;\;tangent\;QMT\;\;and\;\;chord\;NM = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle ONM = \angle OMN = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle NOM \\[3ex] \angle ONM + \angle OMN + \angle NOM = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle NOM \\[3ex] p + p + 54 = 180 \\[3ex] 2p = 180 - 54 \\[3ex] 2p = 126 \\[3ex] p = \dfrac{126}{2} \\[5ex] p = 63 \\[3ex] \therefore \angle ONM = \angle OMN = 63^\circ \\[3ex] \angle OMQ = 90^\circ ...radius\;OM \perp tangent\;QMT\;\;at\;\;point\;\;of\;\;contact\;M \\[3ex] \angle NMQ + \angle OMN = \angle OMQ ...diagram \\[3ex] \angle NMQ + 63 = 90 \\[3ex] \angle NMQ = 90 - 63 \\[3ex] \angle NMQ = 27^\circ $(322.) (323.) NZQA The points P, Q, R, and S all lie on the circumference of circle, with centre C. TP and TQ are tangents to the circle. PCR is a diameter of the circle. Angle PSQ is 68°. RS = 10 cm. PS = 16 cm. (i) Find the size, a, of the angle PRS. Show your working clearly. (ii) Find the size, b, of the angle PTQ. Justify your answer with clear geometric reasoning.$ (i) \\[3ex] \underline{\triangle PRS} \\[3ex] \angle PSR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \triangle PSR \;\;is\;\;a\;\;right-angled\;\;\triangle \\[3ex] \tan \angle PRS = \dfrac{PS}{RS} ...SOHCAHTOA \\[5ex] \tan a = \dfrac{16}{10} \\[5ex] \tan a = 1.6 \\[3ex] a = \tan^{-1}(1.6) \\[3ex] a = 57.99461679 \\[3ex] a \approx 58^\circ \\[5ex] (ii) \\[3ex] \angle PCQ = 2 * \angle PSQ ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle PCQ = 2 * 68 \\[3ex] \angle PCQ = 136^\circ \\[3ex] \angle CQT = \angle CPT = 90^\circ \\[3ex] ...radius\;CQ \perp tangent\;QT\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] ...radius\;CP \perp tangent\;PT\;\;at\;\;point\;\;of\;\;contact\;P \\[3ex] \angle PTQ + \angle CQT + \angle CPT + \angle PCQ = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;CQTP \\[3ex] \implies \\[3ex] b + 90 + 90 + 136 = 360 \\[3ex] b + 316 = 360 \\[3ex] b = 360 - 316 \\[3ex] b = 44^\circ $(324.) GCSE The diagram shows a circle centre O and radius 6cm. The line AB is a tangent to the circle at A. The point C is where the line OB crosses the circumference of the circle. Angle ABO = 34° (a) Explain why the radius OA is the shortest distance from O to the tangent AB. (b) (i) Calculate the length of OB. (ii) Work out the length of BC.$ \underline{\triangle AOB} \\[3ex] \angle AOB + \angle OAB + \angle ABO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle AOB + 90 + 34 = 180 \\[3ex] \angle AOB + 124 = 180 \\[3ex] \angle AOB = 180 - 124 \\[3ex] \angle AOB = 56^\circ \\[3ex] $(a.) Side Length - Angle Measure Theorem: the side length facing an angle is opposite the angle measure as regards size. small angle = ∠ABO = 34° side length facing 34° = radius = |OA| = 6cm Hence, the radius OA is the shortest distance from O to the tangent AB.$ (b) \\[3ex] (i) \\[3ex] \dfrac{|OB|}{\sin 90} = \dfrac{|OA|}{\sin 34}...Sine\;\;Law \\[5ex] |OB| = \dfrac{|OA| * \sin 90}{\sin 34} \\[5ex] |OB| = \dfrac{6 * 1}{0.5591929035} \\[5ex] |OB| = 10.7297499\;cm \\[3ex] (ii) \\[3ex] |OC| = |OA| = 6\;cm \\[3ex] |OC| + |BC| = |OB| ... diagram \\[3ex] 6 + |BC| = 10.7297499 \\[3ex] |BC| = 10.7297499 - 6 \\[3ex] |BC| = 4.729749899\;cm $(325.) (326.) (327.) (328.) (329.) (330.) GCSE ABC is a triangle such that the points A, B and C lie on a circle, centre O. BC = BA and ∠CAB = 32° Find the size, in degrees, of (a) ∠COB (b) ∠OCA$ (a) \\[3ex] \angle COB = 2 * \angle CAB ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle COB = 2 * 32 \\[3ex] \angle COB = 64^\circ \\[3ex] (b) \\[3ex] \angle OCB = \angle OBC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle COB \\[3ex] \angle OCB + \angle OBC + \angle COB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COB \\[3ex] p + p + 64 = 180 \\[3ex] 2p = 180 - 64 \\[3ex] 2p = 116 \\[3ex] p = \dfrac{116}{2} \\[5ex] p = 58^\circ \\[3ex] \therefore \angle OCB = \angle OBC = 58^\circ \[3ex] \angle ACB = \angle CAB = 32^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle CAB \\[3ex] \angle OCA + \angle ACB = \angle OCB ...diagram \\[3ex] \angle OCA + 32 = 58 \\[3ex] \angle OCA = 58 - 32 \\[3ex] \angle OCA = 26^\circ $(331.) NZQA The points F, G, H all lie on the circumference of a semi-circle, with centre C. Angle GFH = x Angle GCH = y Line FCH is straight. Find the size, y, of angle GCH, giving your answer in terms of x Justify your answer with clear geometric reasoning.$ \angle CFG = \angle GFH = x ...diagram \\[3ex] \angle CGF = \angle CFG = x ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle FGC \\[3ex] \angle GCH = \angle CFG + \angle CGF ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] y = x + x \\[3ex] y = 2x $(332.) (333.) (334.) (335.) GCSE A, B and D are three points on a circle. The point C is such that CD and CB are tangents to the circle. ADB = ∠ABD Prove that$\triangle$ABC and$\triangle$ADC are congruent. There are at least four ways to prove that$\triangle ABC$and$\triangle DCB$are congruent. Use any approach you prefer$ \angle ADB = \angle ABD ...given \\[3ex] \implies \\[3ex] |AB| = |AD| ...isosceles\;\triangle \\[3ex] |BC| = |DC| ...Intersecting\;\;Tangents\;\;Theorem \\[3ex] \implies \\[3ex] \angle CBD = \angle CDB ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle BCD \\[3ex] \angle ABC = \angle ABD + \angle CBD ...diagram \\[3ex] \angle ADC = \angle ADB + \angle CDB ...diagram \\[3ex] \implies \\[3ex] \angle ABC = \angle ADC \\[3ex] Also: \\[3ex] \angle BCA = \angle DCA \\[3ex] ... two\;\;tangents:\;BC\;\;and\;\;DC\;\;drawn\;\;from\;\;same\;\;external\;\;point:\;C\;\;bisects\;\;the\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents:\;\angle BCD \\[3ex] |AC| = |AC| ...for\;\;\triangle ABC\;\;and\;\;\triangle ADC \\[3ex] \underline{First\;\;Approach:\;\;SSS} \\[3ex] Side:\;\; |AB| = |AD| \\[3ex] Side:\;\; |BC| = |DC| \\[3ex] Side:\;\; |AC| = |AC| \\[3ex] \underline{Second\;\;Approach:\;\;ASA} \\[3ex] Angle:\;\; \angle ABC = \angle ADC \\[3ex] Side:\;\; |BC| = |DC| \\[3ex] Angle:\;\; \angle BCA = \angle DCA \\[3ex] \underline{Third\;\;Approach:\;\;AAS} \\[3ex] Angle:\;\; \angle ABC = \angle ADC \\[3ex] Angle:\;\; \angle BCA = \angle DCA \\[3ex] Side:\;\; |AC| = |AC| \\[3ex] \underline{Fourth\;\;Approach:\;\;SAS} \\[3ex] Side:\;\; |BC| = |DC| \\[3ex] Angle:\;\; \angle BCA = \angle DCA \\[3ex] Side:\;\; |AC| = |AC| $(336.) (337.) (338.) (339.) NZQA The points E, F, G, and H all lie on the circumference of a circle, C. EH = 8 cm. GH = 12 cm. Angle EHF = 75°. (i) Calculate the length, EG, of the diameter of th circle. Show your working clearly. (ii) Find the size, x, of angle FEH. Justify your answer with clear geometric reasoning.$ (i) \\[3ex] \angle EHG = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \triangle EHG \;\;is\;\;a\;\;right-angled\;\;\triangle \\[3ex] |EG|^2 = |EH|^2 + |GH|^2 ...Pythagorean\;\;Theorem \\[3ex] |EG|^2 = 8^2 + 12^2 \\[3ex] |EG|^2 = 64 + 144 \\[3ex] |EG|^2 = 208 \\[3ex] |EG| = \sqrt{208} \\[3ex] |EG| = 14.4222051 \\[3ex] |EG| \approx 14\;cm \\[5ex] (ii) \\[3ex] \underline{\triangle EHG} \\[3ex] \tan \angle EGH = \dfrac{|EH|}{|GH|} ...SOHCAHTOA \\[5ex] \tan \angle EGH = \dfrac{8}{12} \\[5ex] \tan \angle EGH = 0.6666666667 \\[3ex] \angle EGH = \tan^{-1}(0.6666666667) \\[3ex] \angle EGH = 33.69006753^\circ \\[3ex] \underline{\triangle EFH} \\[3ex] \angle EFH = \angle EGH = 33.69006753^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle FEH + \angle EHF + \angle EFH = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] x + 75 + 33.69006753 = 180 \\[3ex] x + 108.6900675 = 180 \\[3ex] x = 180 - 108.6900675 \\[3ex] x = 71.30993247 \\[3ex] x \approx 71^\circ $(340.) (341.) GCSE A, D, B and E are points on a circle, centre O. AFBC, OEC and OFD are straight lines. AF = 7cm, FB = 4cm, BC = 5cm, FD = 2cm and CE = xcm. Work out the value of x Show your working clearly. Construction: (1.) Draw the diameter by joining the radius from |OD| to the circumference of the circle to G This is to enable us to use the Intersecting Chords theorem so we can find the diameter. (2.) Draw the diameter by joining the radius from |OE| to the circumference of the circle to H. This is to enable us to use the Intersecting Secants theorem so we can find the value of x$ |DG| = diameter = d \\[3ex] |DF| + |FG| = |DG| ...diagram \\[3ex] 2 + |FG| = d \\[3ex] |FG| = d - 2 \\[3ex] |DF| * |FG| = |AF| * |FB| ...Intersecting\;\;Chords\;\;Theorem \\[3ex] 2 * (d - 2) = 7 * 4 \\[3ex] d - 2 = \dfrac{7 * 4}{2} \\[5ex] d - 2 = 14 \\[3ex] d = 14 + 2 \\[3ex] d = 16\;cm \\[3ex] |HE| = diameter = d = 16\;cm \\[3ex] |AB| = |AF| + |FB| ... diagram \\[3ex] |AB| = 7 + 4 = 11\;cm \\[3ex] |EC| * (|HE| + |EC|) = |BC| * (|AB| + |BC|) ...Intersecting\;\;Secants\;\;Theorem \\[3ex] x * (16 + x) = 5(11 + 5) \\[3ex] 16x + x^2 = 5(16) \\[3ex] x^2 + 16x = 80 \\[3ex] x^2 + 16x - 80 = 0 \\[3ex] (x + 20)(x - 4) = 0 \\[3ex] x + 20 = 0 \;\;\;OR\;\;\; x - 4 = 0 \\[3ex] x = -20 \;\;\;OR\;\;\; x = 4 \\[3ex] $Because the length cannot have a negative value:$ x = 4\;cm $(342.) (343.) (344.) SQA National 5 Maths The diagram below shows a circle, centre C. The radius of the circle is 15 centimetres. A is the mid-point of chord PQ. The length of AB is 27 centimetres. Calculate the length of PQ. Construction: Draw and label the radii: from centre C to point P from centre C to point Q$ |CB| = |CP| = |CQ| = radius = r = 15\;cm ...given \\[3ex] |AB| = |AC| + r ...diagram \\[3ex] 27 = |AC| + 15 \\[3ex] |AC| + 15 = 27 \\[3ex] |AC| = 27 - 15 \\[3ex] |AC| = 12\;cm \\[3ex] \underline{\triangle CPA} \\[3ex] |CP|^2 = |PA|^2 + |AC|^2 ...Pythagorean\;\;theorem \\[3ex] 15^2 = |PA|^2 + 12^2 \\[3ex] |PA|^2 + 12^2 = 15^2 \\[3ex] |PA|^2 + 144 = 225 \\[3ex] |PA|^2 = 225 - 144 \\[3ex] |PA|^2 = 81 \\[3ex] |PA| = \sqrt{81} \\[3ex] |PA| = 9\;cm \\[3ex] |AQ| = |PA| = 9\;cm...A\;\;is\;\;the\;\;midpoint\;\;of\;\;chord\;PQ \\[3ex] \implies \\[3ex] |PQ| = |PA| + |AQ| \\[3ex] |PQ| = 9 + 9 \\[3ex] |PQ| = 18\;cm $(345.) (346.) (347.) GCSE A, C and D are points on a circle, centre O. AB and CB are tangents to the circle. Angle ABC = 74° Work out the size of angle ADC Show your working clearly. Construction: To use these theorems: Intersecting Tangents Theorem and Angle of Intersecting Tangents Theorem, we will need to draw the line from the centre of the circle to the intersection of the two tangents: Point B$ \angle OCB = 90^\circ ...radius\;OC \perp tangent\;CB\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] \angle OAB = 90^\circ ...radius\;OA \perp tangent\;AB\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OBC = \angle OBA = \dfrac{\angle ABC}{2} = \dfrac{74}{2} = 37^\circ ... Angle\;\;of\;\;Intersecting\;\;Tangents\;\;Theorem \\[3ex] \angle COB + \angle OBC + \angle OCB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COB \\[3ex] \angle COB + 37 + 90 = 180 \\[3ex] \angle COB + 127 = 180 \\[3ex] \angle COB = 180 - 127 \\[3ex] \angle COB = 53^\circ \\[3ex] \angle AOB = \angle COB = 53^\circ ...Intersecting\;\;Tangents\;\;Theorem \\[3ex] \angle AOC = \angle AOB + \angle COB ...diagram \\[3ex] \angle AOC = 53 + 53 = 106^\circ \\[3ex] \angle AOC = 2 * \angle ADC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 106 = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 106 \\[3ex] \angle ADC = \dfrac{106}{2} \\[5ex] \angle ADC = 53^\circ $(348.) (349.) (350.) (351.) (352.) GCSE A and B are points on the circumference of a circle, centre O CBD is the tangent to the circle at B CAO is a straight line. ABC = 38° Giving your reasons, find the size, in degrees, of ∠ACB$ \angle OBC = 90^\circ \\[3ex] ...radius\;OB \perp tangent\;CBD\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \angle ABC + \angle OBA = \angle OBC ...diagram \\[3ex] 38 + \angle OBA = 90 \\[3ex] \angle OBA = 90 - 38 \\[3ex] \angle OBA = 52^\circ \\[3ex] \angle OAB = \angle OBA = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOB \\[3ex] \angle AOB + \angle OAB + \angle OBA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB + 52 + 52 = 180 \\[3ex] \angle AOB + 104 = 180 \\[3ex] \angle AOB = 180 - 104 \\[3ex] \angle AOB = 76^\circ \\[3ex] \angle CAB = \angle AOB + \angle OBA \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle CAB = 76 + 52 \\[3ex] \angle CAB = 128^\circ $(353.) (354.) (355.) GCSE In the diagram, ABDFE is a circle, centre O and radius 6cm. ABC and EDC are straight lines. The arc EFD is of length$\dfrac{7}{2}\pi$cm. Calculate the length, in cm to 3 significant figures, of DC.$ |OD| = |OE| = radius = r = 6\;cm \\[3ex] Length\;\;of\;\;arc\;EFD = \dfrac{\angle EOD}{360} * 2 * \pi * r \\[5ex] \dfrac{7}{2}\pi = \dfrac{\angle EOD}{360} * 2 * \pi * 6 \\[5ex] \dfrac{7\pi}{2} = \dfrac{\angle EOD}{360} * 12\pi \\[5ex] \dfrac{7}{2} = \dfrac{\angle EOD}{30} \\[5ex] \dfrac{\angle EOD}{30} = \dfrac{7}{2} \\[5ex] \angle EOD = \dfrac{7 * 30}{2} \\[5ex] \angle EOD = 7(15) = 105^\circ \\[3ex] |ED|^2 = |OE|^2 + |OD|^2 - 2 * |OE| * |OD| * \cos \angle EOD...Cosine\;\;Law \\[3ex] |ED|^2 = 6^2 + 6^2 - 2(6)(6) * \cos 105^\circ \\[3ex] = 36 + 36 - 72(-0.2588190451) \\[3ex] = 72 + 18.63497125 \\[3ex] = 90.63497125 \\[3ex] |ED| = \sqrt{90.63497125} \\[3ex] |ED| = 9.520240083\;cm \\[3ex] |DC| * |EC| = |BC| * |AC| ...Intersecting\;\;Secants\;\;Theorem \\[3ex] |EC| = |ED| + |DC| ...diagram \\[3ex] |EC| = 9.520240083 + |DC| \\[3ex] |AC| = |AB| + |BC| ... diagram \\[3ex] |AC| = 7 + 9 = 16\;cm \\[3ex] Let\;\;|DC| = p...to\;\;make\;\;it\;\;simpler \\[3ex] \implies \\[3ex] |DC| * (9.520240083 + |DC|) = 9(16) \\[3ex] p(9.520240083 + p) = 144 \\[3ex] 9.520240083p + p^2 = 144 \\[3ex] p^2 + 9.520240083p - 144 = 0 \\[3ex] Compare\;\;to\;\;general\;\;form:\;\;ap^2 + bp + c = 0 \\[3ex] a = 1 \\[3ex] b = 9.520240083 \\[3ex] c = -144 \\[3ex] p = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} ...Quadratic\;\;Formula \\[5ex] p = \dfrac{-9.520240083 \pm \sqrt{9.520240083^2 - 4(1)(-144)}}{2(1)} \\[5ex] p = \dfrac{-9.520240083 \pm \sqrt{90.63497124 + 576}}{2} \\[5ex] p = \dfrac{-9.520240083 \pm \sqrt{666.6349712}}{2} \\[5ex] p = \dfrac{-9.520240083 \pm 25.81927519}{2} \\[5ex] p = \dfrac{-9.520240083 + 25.81927519}{2} \;\;\;OR\;\;\; p = \dfrac{-9.520240083 - 25.81927519}{2} \\[5ex] p = \dfrac{16.29903511}{2} \;\;\;OR\;\;\; p = \dfrac{-35.33951527}{2} \\[5ex] p = 8.149517553 \;\;\;OR\;\;\; p = -17.66975764 \\[3ex] $Because the length cannot be negative, discard the negative value$ p = 8.149517553 \\[3ex] p \approx 8.15\;cm...to\;\;3\;\;significant\;\;figures $(356.) (357.) (358.) (359.) GCSE Figure 6 shows a triangle ABC and a circle PQR, centre O. The triangle is such that side AB is the tangent to the circle at P, side BC is the tangent to the circle at Q and side AC is the tangent to the circle at R. The region inside the circle is shaded, as shown in Figure 6. AB = 14cm, BC = 12cm and AC = 8cm Let BP = xcm and by considering the lengths of the tangents to the circle, (a) obtain an equation in x only and solve it to find the length, in cm, of BP. (b) Find, to 3 significant figures, the area of the circle as a percentage of the total area of triangle ABC. (a) Construction: To get a clearer picture of x and the remaining lengths, let us indicate the length of x and the radii. Let: |AP| = m |CQ| = n$ |AP| = |AR| = m ...two\;\;tangents:\;\;AP\;\;and\;\;AR\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;A\;\;are\;\;equal\;\;in\;\;length \\[3ex] |CR| = |CQ| = n ...two\;\;tangents:\;\;CR\;\;and\;\;CQ\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] |AR| + |RC| = m + n = 8\;cm...diagram \\[3ex] |BQ| = |BP| = x ...two\;\;tangents:\;\;PB\;\;and\;\;QB\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;B\;\;are\;\;equal\;\;in\;\;length \\[3ex] |AB| + |CB| = 14 + 12 = 26\;cm \\[3ex] \implies \\[3ex] x + m + x + n = |AB| + |CB| \\[3ex] 2x + m + n = 26...eqn.(1) \\[3ex] But: \\[3ex] m + n = 8 ...eqn.(2) \\[3ex] Substitute\;\;eqn.(2)\;\;into\;\;eqn.(1) \\[3ex] 2x + 8 = 26 \\[3ex] 2x = 26 - 8 \\[3ex] 2x = 18 \\[3ex] x = \dfrac{18}{2} \\[5ex] x = 9\;cm \\[3ex] (b) \\[3ex] $Let us determine angle B because we shall need it to answer (b)$ \underline{\triangle ABC} \\[3ex] side = lower\;\;case\;\;letters \\[3ex] angle = upper\;\;case\;\;letters \\[3ex] Let: \\[3ex] |AB| = c = 14\;cm \\[3ex] |AC| = b = 8\;cm \\[3ex] |BC| = a = 12\;cm \\[3ex] b^2 = a^2 + c^2 - 2ac \cos B ...Cosine\;\;Law \\[3ex] 8^2 = 12^2 + 14^2 - 2(12)(14) \cos B \\[3ex] 64 = 144 + 196 - 336\cos B \\[3ex] 64 = 340 - 336\cos B \\[3ex] 336\cos B = 340 - 64 \\[3ex] \cos B = \dfrac{276}{336} \\[5ex] \cos B = 0.8214285714 \\[3ex] B = \cos^{-1}(0.8214285714) \\[3ex] B = 34.77194403^\circ \\[3ex] Area\;\;of\;\;\triangle ABC = \dfrac{1}{2}ac\sin B \\[5ex] = \dfrac{1}{2} * 12 * 14 * \sin 34.77194403^\circ \\[5ex] = 84(0.5703114079) \\[3ex] = 47.90615827\;cm^2 \\[3ex] $We need to find the radius of the circle. So, we can determine the area of the circle. Construction: Draw a line from the centre O to point B In other words, draw line |OB|$ \angle OBP = \angle OBQ = \dfrac{\angle B}{2} ... Angle\;\;of\;\;Intersecting\;\;Tangents\;\;Theorem \\[5ex] \angle OBP = \dfrac{34.77194403}{2} \\[5ex] \angle OBP = 17.38597202^\circ \\[3ex] \underline{\triangle POB} \\[3ex] SOHCAHTOA \\[3ex] \tan \angle OBP = \dfrac{r}{x} \\[5ex] \tan 17.38597202 = \dfrac{r}{9} \\[5ex] 9\tan 17.38597202 = r \\[3ex] r = 9(0.3131121456) \\[3ex] r = 2.818009311\;cm \\[3ex] \underline{Circle\;RPQ} \\[3ex] Area\;\;of\;\;the\;\;circle = \pi r^2 \\[3ex] = \dfrac{22}{7} * (2.818009311)^2 \\[5ex] = 24.95798321\;cm^2 \\[3ex] $What percent of the area of the triangle is the area of the circle?$ What\;\;\%\;\;of\;\;47.90615827\;\;is\;\;24.95798321? \\[3ex] Let\;\;the\;\;variable\% = k \\[3ex] \dfrac{is}{of} = \dfrac{variable}{100} \\[5ex] \dfrac{24.95798321}{47.90615827} = \dfrac{k}{100} \\[5ex] k = \dfrac{24.95798321 * 100}{47.90615827} \\[5ex] k = 52.09765113\% \\[3ex] k \approx 52.1\%...to\;\;3\;\;significant\;\;figures \\[3ex] \$ Approximately 52.1% of the total area of triangle is the area of the circle.
(360.)

(361.)

(362.)