Question

In the figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then $\angle BQD$ =

Answer 1

In the figure, P and Q are the centres of two circles which intersect each other at C and B.

ACD is a straight line.

$\angle APB={150}^o$

Arc AB subtends $\angle APB$ at the centre and $\angle ACB$ at the remaining part of the circle.

$\therefore \angle ACB=\frac{1}{2}\angle APB=\frac{1}{2}\times {150}^o={75}^o$

But $\angle ACB+\angle BCD={180}^o$ (Linear pair)

$\Rightarrow {75}^o+\angle BCD={180}^o$

$\therefore \angle BCD={180}^o-{75}^o={105}^o$

Now, arc BD subtends reflex $\angle BQD$ at the centre and $\angle BCD$ at the remaining part of the circle.

Reflex $\angle BQD=2\angle BCD=2\times {105}^o={210}^o$

But, $\angle BQD+reflex\angle BQD={360}^o$

$\Rightarrow \angle BQD+{210}^o={360}^o$

$\therefore \angle BQD={360}^o-{210}^o={150}^o$