In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, ∠A = 35°, ∠Q = 25°. Find :

(i) ∠PRB

(ii) ∠PBR

(iii) ∠BPR

(i) ∠PRB = ∠PAB = 35° (∵ angles in same segment are equal.)

Hence, the value of ∠PRB = 35°

(ii) From figure,

∠APB = 90° (∵ angle in semicircle is 90°.)

⇒ ∠APB + ∠BPQ = 180° (∵ angles form a linear pair).
⇒ 90° + ∠BPQ = 180°
⇒ ∠BPQ = 90°.

Exterior angle in a triangle is equal to the sum of opposite two interior angles.

In △PBQ,

Ext. ∠PBR = ∠PQB + ∠BPQ = 25° + 90° = 115°.

Hence, the value of ∠PBR = 115°

(iii) In △PRQ,

Ext. ∠APR = ∠PRQ + ∠PQR = ∠PRB + ∠PQR = 35° + 25° = 60°.

∠APB = 90° (∵ angle in semicircle is 90°.)

From figure,

∠BPR = ∠APB - ∠APR = 90° - 60° = 30°.

Hence, the value of ∠BPR = 30°