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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. - Mathematics

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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.

Sum

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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Chapter 17: Circles - Exercise 17 (C) [Page 266]

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Selina Mathematics [English] Class 10 ICSE
Chapter 17 Circles
Exercise 17 (C) | Q 13 | Page 266
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