16. Circles (Angle Properties & Cyclic Quadrilaterals) Previous Year Questions

∠ACB = 180° − 120° = 60°. In ΔABC, AC = BC ⇒ ∠ABC = 60°.
In cyclic quadrilateral ABCE, opposite angles are supplementary: ∠ABC + ∠AEC = 180° ⇒ ∠AEC = 180° − 60° = 120°. [1 Mark]

(a) ∠OQM: 90° − 42° = 48°. [1 Mark]
(b) ∠QNS: In ΔOMS, ∠OMS = 90° − 25° = 65°. ∠QMS = 48° + 65° = 113°. Cyclic quad MQNS ⇒ ∠QNS = 180° − 113° = 67°. [1 Mark]
(c) ∠QOS: 2 × ∠QNS = 2 × 67° = 134°. [1 Mark]
(d) ∠QMS: 113°. [1 Mark]

(i) Arc $AC$ subtends $\angle AOC = 72^\circ$ at the centre and $\angle ABC$ at the circumference.
$\angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 72^\circ = \mathbf{36^\circ}$. (Angle at circumference is half angle at centre). [1 Mark]
(ii) Given chord $AC = \text{chord } BD$. Equal chords subtend equal angles at the centre:
$\angle BOD = \angle AOC = 72^\circ$.
Therefore, $\angle BAD = \frac{1}{2} \angle BOD = \frac{1}{2} \times 72^\circ = \mathbf{36^\circ}$. [1 Mark]
(iii) Since $AB$ is a diameter, $\angle ADB = 90^\circ$ (Angle in a semicircle is a right angle).
In right $\triangle ABD$: $\angle ABD = 180^\circ - (90^\circ + \angle BAD) = 90^\circ - 36^\circ = \mathbf{54^\circ}$. [1 Mark]

Step 1 (i) Find $\angle ACB$:
Since $AC$ is a diameter, the angle subtended by it in the semicircle is a right angle:
$\angle ABC = 90^\circ$ (Angle in a semi-circle).
In $\Delta ABC$: $\angle BAC + \angle ABC + \angle ACB = 180^\circ$
$50^\circ + 90^\circ + \angle ACB = 180^\circ \implies \mathbf{\angle ACB = 40^\circ}$. [1 Mark]
Step 2 (ii) Find $\angle EDC$:
Given $BC \parallel AE$ with transversal $AC$:
$\angle EAC = \angle ACB = 40^\circ$ (Alternate interior angles).
Since vertices $A, C, D, E$ lie on the circle, $ACDE$ is a cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary:
$\angle EAC + \angle EDC = 180^\circ \implies 40^\circ + \angle EDC = 180^\circ \implies \mathbf{\angle EDC = 140^\circ}$. [1 Mark]
Step 3 (iii) Find $\angle BEC$:
Angles subtended by the same chord $BC$ at the circumference are equal:
$\mathbf{\angle BEC = \angle BAC = 50^\circ}$ (Angles in the same segment of a circle). [1 Mark]
Step 4 Prove that $BE$ is also a diameter:
Consider $\angle BAE = \angle BAC + \angle EAC = 50^\circ + 40^\circ = 90^\circ$.
Since chord $BE$ subtends a right angle ($90^\circ$) on the circumference of the circle, by the converse of the semicircle angle theorem, **$BE$ must be a diameter of the circle**.
(Alternatively, in $\Delta BCE$, $\angle BCE = \angle BCA + \angle ACE = 40^\circ + 50^\circ = 90^\circ$, proving $BE$ is a diameter).
Hence Proved. [1 Mark]

Solution In $\Delta PAB$, $PA = PB = AB \implies \Delta PAB$ is equilateral $\implies \angle APB = 60^\circ$.
Central angle $\angle AOB = 180^\circ - \angle APB = 180^\circ - 60^\circ = 120^\circ$.
The angle subtended by arc $AB$ at the circumference is half the central angle:
$\angle ACB = \frac{1}{2}\angle AOB = \frac{1}{2}(120^\circ) = \mathbf{60^\circ}$.
Correct Answer: (c) 60° [1 Mark]

Step 1 (a) Find $\angle PSR$:
Since $PQRS$ is a cyclic quadrilateral, opposite angles are supplementary:
$\angle PSR + \angle PQR = 180^\circ \implies \angle PSR + 112^\circ = 180^\circ \implies \mathbf{\angle PSR = 68^\circ}$. [1 Mark]
Step 2 (b) Find $\angle BPR$:
By Alternate Segment Theorem, the angle between tangent $PB$ and chord $PR$ equals the angle subtended by chord $PR$ in the alternate segment ($\angle PSR$):
$\mathbf{\angle BPR = \angle PSR = 68^\circ}$. [1 Mark]
Step 3 (c) Find $\angle PTS$:
In cyclic quadrilateral $PRST$ or using chord $PS$ with isosceles triangle $PS = SR$:
In $\Delta PSR$, $PS = SR \implies \angle SPR = \angle SRP = \frac{180^\circ - 68^\circ}{2} = 56^\circ$.
Angles subtended by chord $PS$ in the same segment: $\angle PTS = \angle PRS = \mathbf{56^\circ}$. [1 Mark]

(i) Find $\angle QRS$ Opposite angles of a cyclic quadrilateral are supplementary:
$$\angle QPS + \angle QRS = 180^\circ \implies 73^\circ + \angle QRS = 180^\circ$$ $$\angle QRS = 180^\circ - 73^\circ = \mathbf{107^\circ}$$ [1 Mark]
(ii) Find $\angle RQS$ In $\triangle PQS$:
$$\angle PSQ = 180^\circ - (\angle QPS + \angle PQS) = 180^\circ - (73^\circ + 55^\circ) = 180^\circ - 128^\circ = 52^\circ$$ Given $\angle PSR = 82^\circ$, we have:
$$\angle QSR = \angle PSR - \angle PSQ = 82^\circ - 52^\circ = 30^\circ$$ Now in $\triangle QRS$, the angle sum is $180^\circ$:
$$\angle RQS + \angle QSR + \angle QRS = 180^\circ \implies \angle RQS + 30^\circ + 107^\circ = 180^\circ$$ $$\angle RQS = 180^\circ - 137^\circ = \mathbf{43^\circ}$$ [1.5 Marks]
(iii) Find $\angle PRQ$ Angles subtended by the same chord $PQ$ in the same segment of the circle are equal:
$$\angle PRQ = \angle PSQ = \mathbf{52^\circ}$$ [1.5 Marks]
Answers: (i) $\angle QRS = 107^\circ$, (ii) $\angle RQS = 43^\circ$, (iii) $\angle PRQ = 52^\circ$.

Step 1 Since $O$ is the centre, $OP = OQ = \text{radius } r$.
Given $QR = OP$, we have $QR = OQ = r$.
In $\triangle OQR$, $OQ = QR \implies \triangle OQR$ is isosceles.
$$\angle QOR = \angle QRO = 20^\circ$$ [1 Mark]
Step 2 In $\triangle OQR$, exterior angle at $Q$ is $\angle OQP$:
$$\angle OQP = \angle QOR + \angle QRO = 20^\circ + 20^\circ = 40^\circ$$ In $\triangle OPQ$, $OP = OQ = r \implies \triangle OPQ$ is also isosceles.
$$\angle OPQ = \angle OQP = 40^\circ$$ [1 Mark]
Step 3 For $\triangle OPR$, the exterior angle at vertex $O$ on straight line $TOR$ is $\angle POT = x$:
$$x = \angle OPR + \angle ORP = 40^\circ + 20^\circ = \mathbf{60^\circ}$$ [1 Mark]
Alternative: In $\triangle OPQ$, $\angle POQ = 180^\circ - (40^\circ + 40^\circ) = 100^\circ$.
On straight line $TOR$: $x + \angle POQ + \angle QOR = 180^\circ \implies x + 100^\circ + 20^\circ = 180^\circ \implies x = 60^\circ$.
Answer: $x = 60^\circ$.

Step 1 The perpendicular from the centre of a circle to a chord bisects the chord.
For chord $AB = 24\text{ cm}$: $AN = \frac{24}{2} = 12\text{ cm}$.
In right $\triangle ONA$ ($OA = 13\text{ cm}$):
$$ON = \sqrt{OA^2 - AN^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = \mathbf{5\text{ cm}}$$ [1 Mark]
Step 2 For chord $CD = 10\text{ cm}$: $CM = \frac{10}{2} = 5\text{ cm}$.
In right $\triangle OMC$ ($OC = 13\text{ cm}$):
$$OM = \sqrt{OC^2 - CM^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = \mathbf{12\text{ cm}}$$ [1 Mark]
Step 3 Since the chords lie on opposite sides of the centre $O$:
$$\text{Distance between chords } NM = ON + OM = 5 + 12 = \mathbf{17\text{ cm}}$$ [1 Mark]
Answer: Distance between the two chords $= 17\text{ cm}$.

(i) Find ∠BCD In cyclic quadrilateral $ABCD$, the exterior angle is equal to the interior opposite angle:
$$\angle BCD = \angle DAE = \mathbf{70^\circ}$$ [1.5 Marks]
(ii) Find ∠BOD The angle subtended by arc $BCD$ at the centre is twice the angle subtended by it at the circumference:
$$\angle BOD = 2 \times \angle BCD = 2 \times 70^\circ = \mathbf{140^\circ}$$ [1.5 Marks]
(iii) Find ∠OBD In $\triangle OBD$, $OB = OD = \text{radius } r$, so $\triangle OBD$ is isosceles:
$$\angle OBD = \angle ODB = \frac{180^\circ - \angle BOD}{2} = \frac{180^\circ - 140^\circ}{2} = \mathbf{20^\circ}$$ [1 Mark]
Answers: (i) $\angle BCD = 70^\circ$, (ii) $\angle BOD = 140^\circ$, (iii) $\angle OBD = 20^\circ$.