Question

In the following figure, O is the centre of the circle, ∠ PBA = 42°.
Calculate:
(i) ∠ APB
(ii) ∠PQB
(iii) ∠ AQB

Answer 1

(i) In circle C(O, r)
AB is the diameter.
So, ∠ APB = 90°         ....(Angle in a semicircle)

(ii) Now in Δ APB,
∠ PAB = 180° - (∠ APB + ∠ ABP)
∠ PAB = 180° - ( 90° + 42°)
∠ PAB = 180° - 132°
∠ PAB = 48°

∠ PQB = ∠ PAB = 48°     ....(Angle of the same segment)
Hence,
∠ PQB = 48° 

(iii) AQBP is a cyclic quadrilateral.
∠ APB + ∠ AQB = 180°
⇒ 90° + ∠ AQB = 180°
⇒ ∠ AQB = 180° - 90° = 90°.