#### 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°, = 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 + ∠ = 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°