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Solutions
Given:
PQRS is a cyclic quadrilateral inscribed in a circle with centre O.
∠PQR : ∠PSR = 11 : 7
PQ = QR (i.e., triangle PQR is isosceles)
Concept Used:
Opposite angles in a cyclic quadrilateral are supplementary:
∠PQR + ∠PSR = 180°
In an isosceles triangle (PQ = QR), base angles are equal:
∠QPR = ∠PRQ
In any triangle, the sum of all angles is 180°.
The angle subtended by a chord at the centre is twice the angle subtended at the circumference.
Formula Used:
Opposite angles in cyclic quad: ∠PQR + ∠PSR = 180°
Base angles in isosceles triangle: ∠QPR = ∠PRQ
Central angle theorem: ∠POQ = 2 × ∠PRQ

Let ∠PQR = 11x and ∠PSR = 7x
=> 11x + 7x = 180°
=> 18x = 180°
=> x = 10°
Thus, ∠PQR = 110°, ∠PSR = 70°
In triangle PQR:
PQ = QR => ∠QPR = ∠PRQ = y (say)
∠PQR = 110°, so
y + y + 110° = 180°
=> 2y = 70°
=> y = 35°
Then, ∠POQ = 2 × ∠PRQ = 2 × 35° = 70°
Now in triangle OPQ:
OP = OQ => triangle is isosceles
∠POQ = 70°, so remaining 2 angles are equal and each is:
