Mathematics
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.
Circles
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Answer
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 = = 30°.
∠A = 2x = 2 x 30° = 60°
∠B = x = 30°.
Hence, ∠A = 60° and ∠B = 30°.
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