In a cyclic quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A whereas sides PQ and SR produced meet at point B.

If ∠A : ∠B = 2 : 1, find angles A and B.

Let ∠A = 2x and ∠B = x.

In cyclic quadrilateral PQRS,

⇒ ∠PSR + ∠PQR = 180° [As sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠PSR + 135° = 180°

⇒ ∠PSR = 180° - 135° = 45°.

From figure,

⇒ ∠PQR + ∠PQA = 180° [Linear pairs]

⇒ ∠PQA + 135° = 180°

⇒ ∠PQA = 180° - 135° = 45°.

In △PBS,

⇒ ∠BSP + ∠BPS + ∠PBS = 180°

⇒ ∠PSR + ∠BPS + ∠B = 180°

⇒ ∠BPS + 45° + x = 180°

⇒ ∠BPS = 180° - 45° - x

⇒ ∠BPS = 135° - x …………(1)

We know that,

An exterior angle is equal to the sum of two opposite interior angles.

In △PQA,

⇒ ∠BPS = ∠PQA + ∠A

⇒ ∠BPS = 45° + 2x ………..(2)

From 1 and 2 we get,

⇒ 135° - x = 45° + 2x

⇒ 2x + x = 135° - 45°

⇒ 3x = 90°

⇒ x = $\dfrac{90°}{3}$ = 30°.

∠A = 2x = 2 x 30° = 60°

∠B = x = 30°.

Hence, ∠A = 60° and ∠B = 30°.