In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.
(a) 72°
(b) 18°
(c) 36°
(d) 54°

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let $\angle AQP=\angle BQR=x°$
Now,
$$\begin{gathered} \angle AQP+\angle PQR+\angle BQR=180°\left[\text{∵}\mathit{\text{AQB}}\text{is a straight line}\right] \\ \Rightarrow x+108+x=180° \\ \Rightarrow 2x=72° \\ \Rightarrow x=36° \\ \therefore \angle AQP=36° \end{gathered}$$