Question

In the given figure; AB is a mirror, PQ is the incident ray and QR is the reflected ray. If $\angle PQR={108}^{\circ }$ then $\angle AQP=?$

$\left(a\right){72}^{\circ }$

$\left(b\right){18}^{\circ }$

$\left(c\right){36}^{\circ }$

$\left(d\right){54}^{\circ }$

Answer 1

ANSWER:
$\left(c\right){36}^{\circ }$
We know that angle of incidence = angle of reflection.
Then, let $\angle AQP=\angle BQR=x$
Now,
$\angle AQP+\angle PQR+\angle BQR={180}^{\circ }\because AQBisastraightline\Rightarrow x+108+x={180}^{\circ }\Rightarrow 2x=72°\Rightarrow x={36}^{\circ }\therefore \angle AQP={36}^{\circ }$