In the figure given below, P and Q are centres of two circles, interse

1.	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» 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.

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»

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.

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