Question

In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line. If ∠APB = 150° and ∠BQD = x°, find the value of x.

Answer 1

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

∠APB = 2 ∠ACB

It can be written as

∠ACB = ½ ∠APB

By substituting the values

∠ACB = 150/2

So we get

∠ACB = 75^o

We know that ACD is a straight line

It can be written as

∠ACB + ∠DCB = 180^o

By substituting the values

75^o + ∠DCB = 180^o

On further calculation

∠DCB = 180^o – 75^o

By subtraction

∠DCB = 105^o

We know that

∠DCB = ½ × reflex ∠BQD

By substituting the values

105^o = ½ × (360^o – x)

On further calculation

210^o = 36^o – x

By subtraction

x = 150^o

Therefore, the value of x is 150^o.