In the figure given below, P and Q are centres of two circles, intersecting at B and C,

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.

 

Solution:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part  of the circle.

Here, arc AEB subtends ∠APB at the centre and ∠ACB at C on the circle.

∴ ∠APB = 2∠ACB

$\Rightarrow \angle A C B=\frac{150^{\circ}}{2}=75^{\circ}$          ...(1)

Since ACD is a straight line, ∠ACB + BCD = 180
⇒ ∠BCD = 180 − 75
⇒ ∠BCD = 105            ...(2)

Also, arc BFD subtends reflex ∠BQD at the centre and ∠BCD at C on the circle.

∴ reflex ∠BQD = 2∠BCD

$\Rightarrow$ reflex $\angle B Q D=2\left(105^{\circ}\right)=210^{\circ}$      ...(3)

Now,
reflex ∠BQD + ∠BQD = 360
⇒ 210 + x = 360
⇒ x = 360 − 210
⇒ x = 150

Hence, x = 150.

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