Question

S
11 In the figure, O is the centre of the circle.
BP is parallel to OQ. If PÊA = 42°, calculate
QPA, BPC nd OỘP.
Р
B
O​

Answer 1

Answer:

(i) In circle C(O, r)

AB is the diameter.

So ∠APB = 90˚ (Angle in semi–circle)

(ii) Now in △APB

∠PAB = 180˚ – (∠APB + ∠ABP)

= 180˚ – (90˚ + 42˚)

= 180˚ – 132˚ = 48˚

∠PQB = ∠PAB = 48˚ (Angles of the same segment )

Hence

∠PQB = 48˚