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 Dominic Chukwuemeka

# Solved Examples - Circle Theorems 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.

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

(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.

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

$(i) \\[3ex] \angle ABC = 90^\circ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle ABC = \angle DBA + \angle DBC ...as\:\:shown \\[3ex] 90 = \angle DBA + 46 \\[3ex] \angle DBA + 46 = 90 \\[3ex] \angle DBA = 90 - 46 \\[3ex] \angle DBA = 44^\circ \\[3ex] (ii) \\[3ex] \angle DAC = 46^\circ ... \angle s \:\:in\:\:the\:\:same\:\:segment \\[3ex] (iii) \\[3ex] \angle BCE = \angle A ... exterior \angle \:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] \angle A = \angle DAC + \angle CAB ...as\:\:shown \\[3ex] \angle A = 46 + 28 \\[3ex] \angle A = 74 \\[3ex] \therefore \angle BCE = 74^\circ$
(2.) JAMB If $O$ is the center of the circle in the figure above, find the value of $x$

$A.\:\: 50^\circ \\[3ex] B.\:\: 260^\circ \\[3ex] C.\:\: 100^\circ \\[3ex] D.\:\: 65^\circ \\[3ex] E.\:\: 130^\circ \\[3ex]$

$y = 2(130) ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] y = 260 \\[3ex] \angle O = x + y = 360 ...\angle s \:\:around\:\:a\:\:point \\[3ex] x = 360 - y \\[3ex] x = 360 - 260 \\[3ex] x = 100^\circ$
(3.) WASSCE The diagram is a circle with centre $O$
$U, V, W$ and $Y$ are points on the circle.
Find $\angle UVW$

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

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

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

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

$Let\:\: \angle OPK = x \\[3ex] \angle OPR = \angle ORP = y ...base\:\:\angle s \:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OPR} \\[3ex] y + y + 108 = 180 ... sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 2y = 180 - 108 \\[3ex] 2y = 72 \\[3ex] y = \dfrac{72}{2} \\[5ex] y = 36 \\[3ex] \angle POR = 2 * \angle PKR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PKR \\[3ex] \angle PKR = \dfrac{108}{2} \\[5ex] \angle PKR = 54 \\[3ex] \underline{\triangle PKR} \\[3ex] \angle KPR + \angle KRP + \angle PKR = 180 ... sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle KPR = \angle OPK + \angle OPR \\[3ex] \angle KPR = x + 36 \\[3ex] \angle KRP = 20 + 36 = 56 \\[3ex] \therefore x + 36 + 56 + 54 = 180 \\[3ex] x + 146 = 180 \\[3ex] x = 180 - 146 \\[3ex] x = 34^\circ$
(6.) WASSCE Find the value of $y$ in the diagram.

$A.\:\: 10 \\[3ex] B.\:\: 15 \\[3ex] C.\:\: 20 \\[3ex] D.\:\: 30 \\[3ex]$

$x = 2y ...\angle s \:\:in\:\:the\:\:same\:\:segment \\[3ex] p + 60 = 180 ...\angle s \:\:in\:\:a\:\:straight\:\:line \\[3ex] p = 180 - 60 \\[3ex] p = 120 \\[3ex] 120 + 2y + 2x = 180 ... sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 120 + x + 2x = 180 \\[3ex] 3x = 180 - 120 \\[3ex] 3x = 60 \\[3ex] x = \dfrac{60}{3} \\[5ex] x = 20 \\[3ex] x = 2y \rightarrow y = \dfrac{x}{2} \\[5ex] y = \dfrac{20}{2} \\[5ex] y = 10^\circ$
(7.) CSEC The diagram below, not drawn to scale, shows a circle with center $O$.
$HJ$ 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:

$(i)\:\: \angle OJH \\[3ex] (ii)\:\: \angle JOG \\[3ex] (iii)\:\: \angle JKG \\[3ex] (iv)\:\: \angle JLG \\[3ex]$

$(i) \\[3ex] \angle OJH = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] (ii) \\[3ex] \angle OGH = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \underline{Quadrilateral\:\:JOGH} \\[3ex] \angle J + \angle O + \angle G + \angle H = 360 ...sum\:\:of\:\:interior\:\:\angle s\:\:of\:\:a\:\:Quad \\[3ex] 90 + \angle O + 90 + 48 = 360 \\[3ex] 228 + \angle O = 360 \\[3ex] \angle O = 360 - 228 \\[3ex] \angle O = 132 \\[3ex] \angle O = \angle JOG = 132^\circ \\[3ex] (iii) \\[3ex] \angle JOG = 2 * \angle JKG ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 132 = 2 * \angle JKG \\[3ex] \angle JKG = \dfrac{132}{2} \\[5ex] \angle JKG = 66^\circ \\[3ex] (iv) \\[3ex] Reflex\angle JOG + Obtuse\angle JOG = 360 ...\angle s\:\:around\:\:a\:\:point \\[3ex] Reflex\angle JOG + 132 = 360 \\[3ex] Reflex\angle JOG = 360 - 132 \\[3ex] Reflex\angle JOG = 228 \\[3ex] Reflex\angle JOG = 2 * \angle JLG ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 228 = 2 * \angle JLG \\[3ex] \angle JLG = \dfrac{228}{2} \\[5ex] \angle JLG = 114^\circ \\[3ex] OR \\[3ex] \underline{Quadrilateral\:\:KJLG} \\[3ex] \angle K + \angle L = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] 66 + \angle L = 180 \\[3ex] \angle L = 180 - 66 \\[3ex] \angle L = 114 \\[3ex] \angle L = \angle JLG = 114^\circ$
(8.) WASSCE In the diagram, $O$ is the center of the circle $PQRS$ and $\angle PQR = 65^\circ$.
Find the value of the angle $x$

$A.\:\: 115^\circ \\[3ex] B.\:\: 130^\circ \\[3ex] C.\:\: 230^\circ \\[3ex] D.\:\: 255^\circ \\[3ex]$

$Obtuse\angle O = 2(65) ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] Obtuse\angle O = 130 \\[3ex] Reflex\angle ROP + Obtuse\angle ROP = 360 ...\angle s\:\:around\:\:a\:\:point \\[3ex] Reflex\angle ROP = x \\[3ex] Obtuse\angle ROP = 130 \\[3ex] x + 130 = 360 \\[3ex] x = 360 - 130 \\[3ex] x = 230^\circ$
(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 $\angle QRS$

$\angle SQR = 79^\circ \\[3ex] \angle QRS = x \\[3ex] Reflex\:\:P\hat{O}S = 252^\circ \\[3ex] Reflex\:\:P\hat{O}S + Obtuse\:\:P\hat{O}S = 360 ...\angle s\:\:around\:\:a\:\:point \\[3ex] Obtuse\:\:P\hat{O}S = 360 - 252 \\[3ex] Obtuse\:\:P\hat{O}S = 108 \\[3ex] \underline{\triangle POS} \\[3ex] \angle SPO = \angle PSO = y...base\:\:\angle s\:\:of\:\:isosceles\:\:\triangle \\[3ex] \angle SPO + \angle POS + \angle PSO = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] x + 108 + x = 180 \\[3ex] 2x + 180 - 108 \\[3ex] 2x = 72 \\[3ex] x = \dfrac{72}{2} \\[5ex] x = 36 \\[3ex] Obtuse\:\:P\hat{O}S = 2 * \angle PQS ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PQS \\[3ex] \angle PQS = \dfrac{108}{2} \\[5ex] \angle PQS = 54 \\[3ex] \underline{\triangle PQS} \\[3ex] \angle QPS = \angle QSP = p...base\:\:\angle s\:\:of\:\:isosceles\:\:\triangle \\[3ex] \angle QPS + \angle QSP + \angle PQS = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] p + p + 54 = 180 \\[3ex] 2p + 54 = 180 \\[3ex] 2p = 180 - 54 \\[3ex] 2p = 126 \\[3ex] p = \dfrac{126}{2} \\[5ex] p = 63 \\[3ex] \angle QSR = 63 ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] \\[3ex] OR \\[3ex] \angle PSQ = \angle PSO + \angle OSQ ...as\:\:shown \\[3ex] 63 = 36 + \angle OSQ \\[3ex] \angle OSQ = 63 - 36 \\[3ex] \angle OSQ = 27 \\[3ex] \angle OSR = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \angle OSR = \angle OSQ + \angle QSR ...as\:\:shown \\[3ex] 90 = 27 + \angle QSR \\[3ex] \angle QSR = 90 - 27 \\[3ex] \angle QSR = 63 \\[3ex] \underline{\triangle QSR} \\[3ex] \angle QRS + \angle QRS + \angle SQR = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 63 + x + 79 = 180 \\[3ex] 142 + x = 180 \\[3ex] x = 180 - 142 \\[3ex] x = 38^\circ$
(10.) JAMB In the figure above, $PT$ is a tangent to the circle with centre $O$.
If $PQT = 30^\circ$, find the value of $PTO$

$A.\:\: 30^\circ \\[3ex] B.\:\: 15^\circ \\[3ex] C.\:\: 24^\circ \\[3ex] D.\:\: 12^\circ \\[3ex] E.\:\: 60^\circ \\[3ex]$

$\angle OPT = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \underline{\triangle QPT} \\[3ex] 30 + x + 90 + 2x + x = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 4x + 120 = 180 \\[3ex] 4x = 180 - 120 \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15 \\[3ex] \angle PTO = 2x \\[3ex] \angle PTO = 2(15) \\[3ex] \angle PTO = 30^\circ$
(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 $QRS$? $A.\:\: 50^\circ \\[3ex] B.\:\: 40^\circ \\[3ex] C.\:\: 110^\circ \\[3ex] D.\:\: 80^\circ \\[3ex] E.\:\: 100^\circ \\[3ex]$

$\underline{\triangle PSQ} \\[3ex] \angle PSQ = 50^\circ ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle PSQ = \angle PQS = 50 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \angle PSQ + \angle PQS + \angle SPQ = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 50 + 50 + \angle SPQ = 180 \\[3ex] 100 + \angle SPQ = 180 \\[3ex] \angle SPQ = 180 - 100 \\[3ex] \angle SPQ = 80 \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:SPQR} \\[3ex] \angle P + \angle R = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] 80 + \angle R = 180 \\[3ex] \angle R = 180 - 80 \\[3ex] \angle R = \angle QRS = 100^\circ$
(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 $\angle ABC$

$\angle OCA = \angle OAC = 48^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \angle ACB = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle CAB = \angle OCA = 48 ...as\:\:shown \\[3ex] \rightarrow 48 + \angle ABC + 90 = 180 \\[3ex] 138 + \angle ABC = 180 \\[3ex] \angle ABC = 180 - 138 \\[3ex] \angle ABC = 42^\circ$
(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

$(i)\:\: H\hat{K}L \\[3ex] (ii)\:\: J\hat{O}K \\[3ex] (iii)\:\: J\hat{H}K \\[3ex]$

$(i) \\[3ex] \angle HJK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle HJK = \angle HJL + \angle LJK...as\:\:shown \\[3ex] 90 = 20 + \angle LJK \\[3ex] \angle LJK = 90 - 20 \\[3ex] \angle LJK = 70 \\[3ex] \angle LHK = \angle LJK ...\angle s \:\:in\:\:a\:\:straight\:\:line \\[3ex] \rightarrow \angle LHK = 70 \\[3ex] \angle HLK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle HKL} \\[3ex] \angle LHK + \angle HLK + \angle HKL = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 70 + 90 + \angle HKL = 180 \\[3ex] 160 + \angle HKL = 180 \\[3ex] \angle HKL = 180 - 160 \\[3ex] \angle HKL = 20^\circ \\[3ex] (ii) \\[3ex] \angle OKJ = \angle OJK ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \rightarrow \angle OJK = 50 \\[3ex] \underline{\triangle JOK} \\[3ex] \angle OKJ + \angle OJK + \angle JOK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 50 + 50 + \angle JOK = 180 \\[3ex] 100 + \angle JOK = 180 \\[3ex] \angle JOK = 180 - 100 \\[3ex] \angle JOK = 80^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle JHK} \\[3ex] \angle HJK + \angle JKH + \angle JHK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 90 + 50 + \angle JHK = 180 \\[3ex] 140 + \angle JHK = 180 \\[3ex] \angle JHK = 180 - 140 \\[3ex] \angle JHK = 40^\circ$
(14.) WASSCE Find the value of $\angle TSP$ in the diagram.

$A.\:\: 91^\circ \\[3ex] B.\:\: 89^\circ \\[3ex] C.\:\: 71^\circ \\[3ex] D.\:\: 69^\circ \\[3ex]$

$\angle TSP = \angle TQP ... \angle s \:\:in\:\:the\:\:same\:\:segment \\[3ex] \angle TQP = 31 + 58 ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle TQP = 89 \\[3ex] \therefore \angle TSP = 89^\circ$
(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.

$A.\:\: 40^\circ \\[3ex] B.\:\: 36^\circ \\[3ex] C.\:\: 20^\circ \\[3ex] D.\:\: 9^\circ \\[3ex]$

$x = y + 17...eqn.(1) ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] 2x - 43 = y...eqn.(2) ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] Substitute\:\:eqn.(1)\:\:into\:\:eqn.(2) \\[3ex] 2(y + 17) - 43 = y \\[3ex] 2y + 34 - 43 = y \\[3ex] 2y - y = 43 - 34 \\[3ex] y = 9^\circ$
(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 $x$

$A.\:\: 34^\circ \\[3ex] B.\:\: 56^\circ \\[3ex] C.\:\: 68^\circ \\[3ex] D.\:\: 112^\circ \\[3ex]$

$\angle ORP = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \angle ORP = \angle ORQ + \angle QRP \\[3ex] 90 = \angle ORQ + 34 \\[3ex] \angle ORQ = 90 - 34 \\[3ex] \angle ORQ = 56 \\[3ex] \angle ORQ = \angle OQR = 56 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle ORQ} \\[3ex] \angle ORG + \angle OQR + x = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 56 + 56 + x = 180 \\[3ex] 112 + x = 180 \\[3ex] x = 180 - 112 \\[3ex] x = 68^\circ$
(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 \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \angle ODA = \angle OAD = y ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OAD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[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 ...\angle s\:\:in\:\: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 ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \rightarrow \angle ACD = 57^\circ \\[3ex] (ii) \\[3ex] \angle EAD = \angle ACD ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \rightarrow \angle EAD = 57^\circ \\[3ex] \underline{\triangle AED} \\[3ex] \angle ADE + \angle EAD + \angle AED = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[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 ...as\:\:shown \\[3ex] \angle CDG = \angle CAD ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] \therefore \angle CAD = 18 \\[3ex] \rightarrow 33 = \angle OAC + 18 \\[3ex] \angle OAC = 33 - 18 \\[3ex] \angle OAC = 15^\circ \\[3ex] (iv) \\[3ex] \underline{Quadrilateral\:\:ABCD} \\[3ex] \angle B + \angle D = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] \angle D = \angle ODA + \angle ODC ...as\:\:shown \\[3ex] \angle ODG = \angle ODC + \angle CDG ...as\:\:shown \\[3ex] 90 = \angle ODC + 18 \\[3ex] \angle ODC = 90 - 18 \\[3ex] \angle ODC = 72 \\[3ex] \rightarrow \angle D = 33 + 72 \\[3ex] \angle D = 105 \\[3ex] \rightarrow \angle B + 105 = 180 \\[3ex] \angle B = 180 - 105 \\[3ex] \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 $\angle QMP$

$A.\:\: 138^\circ \\[3ex] B.\:\: 132^\circ \\[3ex] C.\:\: 42^\circ \\[3ex] D.\:\: 48^\circ \\[3ex]$

$\angle MQP = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle MPQ = \angle MNQ ...\angle s\:\:in\:\:same\:\:segment \\[3ex] \rightarrow \angle MPQ = 42 \\[3ex] \underline{\triangle QMP} \\[3ex] \angle QMP + \angle MPQ + \angle MQP = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle QMP + 42 + 90 = 180 \\[3ex] \angle QMP + 132 = 180 \\[3ex] \angle QMP = 180 - 132 \\[3ex] \angle QMP = 48^\circ$
(19.) WASSCE In the diagram, $O$ is the center of the circle.
$\angle OQR = 32^\circ$ and $\angle TPQ = 15^\circ$
Calculate:

$(i)\:\: \angle QPR \\[3ex] (ii)\:\: \angle TQO \\[3ex]$

$(i) \\[3ex] \angle OQR = \angle ORQ = 32 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OQR} \\[3ex] \angle OQR + \angle ORQ + \angle QOR = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 32 + 32 + \angle QOR = 180 \\[3ex] 64 + \angle QOR = 180 \\[3ex] \angle QOR = 180 - 64 \\[3ex] \angle QOR = 116 \\[3ex] \angle QOR = 2 * \angle QPR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 116 = 2 * \angle QPR \\[3ex] \angle QPR = \dfrac{116}{2} \\[5ex] \angle QPR = 58^\circ \\[3ex] (ii) \\[3ex] \underline{Quadrilateral\:\:TPRQ} \\[3ex] \angle P + \angle Q = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] \angle P = \angle TPQ + \angle QPR ...as\:\:shown \\[3ex] \angle P = 15 + 58 \\[3ex] \angle P = 73 \\[3ex] \angle Q = \angle TQO + \angle OQR ...as\:\:shown \\[3ex] \angle Q = \angle TQO + 32 \\[3ex] \rightarrow 73 + \angle TQO + 32 = 180 \\[3ex] \angle TQO + 105 = 180 \\[3ex] \angle TQO = 180 - 105 \\[3ex] \angle TQO = 75^\circ$
(20.) JAMB From the cyclic quadrilateral $TUVW$ above, find the value of $x$

$A.\:\: 24^\circ \\[3ex] B.\:\: 26^\circ \\[3ex] C.\:\: 20^\circ \\[3ex] D.\:\: 23^\circ \\[3ex]$

$\underline{Quadrilateral\:\:TUVW} \\[3ex] \angle U + \angle W = 180 ...sum\:\:of\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] 3x + 20 + 88 = 180 \\[3ex] 3x + 108 = 180 \\[3ex] 3x = 180 - 108 \\[3ex] 3x = 72 \\[3ex] x = \dfrac{72}{3} \\[5ex] x = 24^\circ$

(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 $V$
$VWX$ 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.

$a)\:\: V\hat{Z}W \\[3ex] b)\:\: X\hat{Y}Z \\[3ex]$

$a) \\[3ex] \angle VZW = 51^\circ ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] b) \\[3ex] \angle VWZ = 78^\circ ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:WXYZ} \\[3ex] \angle XYZ = 78^\circ ...exterior\:\: \angle \:\:of\:\:a\:\:cyclic\:\:Quad = interior\:\:opposite\:\: \angle$
(22.) WASSCE In the diagram, $\overline{RQ}$ is a tangent to the circle, $|PS| = |SQ|$ and $\angle SQR = 50^\circ$
Calculate $\angle SRQ$

$A.\:\: 30^\circ \\[3ex] B.\:\: 35^\circ \\[3ex] C.\:\: 40^\circ \\[3ex] D.\:\: 45^\circ \\[3ex]$

$\angle SRQ = \angle SPQ ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \rightarrow \angle SPQ = 50 \\[3ex] \angle SPQ = \angle SQP ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \rightarrow \angle SQP = 50 \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ = \angle SPQ + \angle SQP ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle RSQ = 50 + 50 \\[3ex] \angle RSQ = 100 \\[3ex] \angle SRQ + \angle RSQ + \angle SQR = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle SRQ + 100 + 50 = 180 \\[3ex] \angle SRQ + 150 = 180 \\[3ex] \angle SRQ = 180 - 150 \\[3ex] \angle SRQ = 30^\circ$
(23.) 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 $\angle TKC$

$\angle TAC = 30 ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle ATC = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle TKC} \\[3ex] \angle TCK = 90 + 30 ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle TCK = 120 \\[3ex] \angle TKC + \angle TCK + \angle CTK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle TKC + 120 + 30 = 180 \\[3ex] \angle TKC + 150 = 180 \\[3ex] \angle TKC = 180 - 150 \\[3ex] \angle TKC = 30^\circ$
(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.

$(i)\:\: \angle ACB \\[3ex] (ii)\:\: \angle CBD \\[3ex] (iii)\:\: \angle AED \\[3ex]$

$(i) \\[3ex] \angle BOA = 2 * \angle ACB ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 130 = 2 * \angle ACB \\[3ex] \angle ACB = \dfrac{130}{2} \\[5ex] \angle ACB = 65^\circ \\[3ex] (ii) \\[3ex] \angle CBD = \angle CAD ...\angle s\:\:in\:\:same\:\:segment \\[3ex] \therefore \angle CBD = 30^\circ \\[3ex] (iii) \\[3ex] \angle ADB = \angle ACB ...\angle s\:\:in\:\:same\:\:segment \\[3ex] \rightarrow \angle ADB = 65 \\[3ex] \underline{\triangle AED} \\[3ex] \angle DAE + \angle AED + \angle EDA = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle EDA = \angle ADB = 65 ...as\:\:shown \\[3ex] 30 + \angle AED + 65 = 180 \\[3ex] \angle AED + 95 = 180 \\[3ex] \angle AED = 180 - 95 \\[3ex] \angle AED = 85^\circ$