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Question
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.
Solution
PQRS is a cyclic quadrilateral in which ∠PQR = 135°
Sides SP and RQ are produced to meet at A and sides PQ and SR are produced to meet at B.
∠A : ∠B = 2 : 1
Let ∠A = 2x, then ∠B = x
Now, in cyclic quad PQRS,
Since, ∠PQR = 135°, ∠S = 180° – 135° = 45°
[Since sum of opposite angles of a cyclic quadrilateral are supplementary]
Since, ∠PQR and ∠PQA are linear pair,
∠PQR + ∠PQA = 180°
`=>` 135° + ∠PQA = 180°
`=>` ∠PQA = 180° – 135° = 45°
Now, In ∆PBS,
∠P = 180° – (45° + x)
= 180° – 45° – x
= 135° – x ...(1)
Again, in ∆PQA,
EXT ∠P = ∠PQA + ∠A
= 45° + 2x ...(2)
From (1) and (2),
45° + 2x = 135° – x
`=>` 2x + x = 135° – 45°
`=>` 3x = 90°
`=>` x = 30°
Hence, ∠A = 2x = 2 × 30° = 60° and ∠B = x = 30°
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