Question

AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines.

Find :

(i) $\angle$ PRB,

(ii) $\angle$ PBR,

(iii) $\angle$ BPR.

Answer 1

i)Given, ∠A = 35°, ∠Q = 25°

Now, ∠PAB = ∠PRB = 35° [Angles in the same segment of a circle are equal]

iii)Also, ∠APB = 90° [Angle in a semi-circle is always a right angle]

⇒ ∠APB + ∠BPQ = 180° [Linear pair]

⇒ ∠BPQ = 180° - 90° = 90°

Now, in triangle PQR, by angle sum property of triangle, we have

∠PQR + ∠RPQ + ∠PRQ = 180°

⇒ ∠RPQ = 180° - 35° - 25° = 120°

ii)Therefore, ∠BPR = ∠RPQ - ∠BPQ = 120° - 90° = 30°

Again, in triangle PBR by angle sum property of triangle, we have

∠PBR + ∠R + ∠BPR = 180°

⇒ ∠PBR = 180° - (∠R + ∠BPR)

= 180° - 35° - 30°

= 115°