Question

In the given figure, AB is a mirror; PQ is the incident ray and QR, the reflected ray. If ∠PQR = 112°, find ∠PQA.

Answer 1

We know that the angle of incidence = angle of reflection.
Hence, let $\angle PQA=\angle BQR=x°$
Since, AQB is a straight line, we have:
$$\begin{gathered} \therefore \angle PQA+\angle PQR+\angle BQR=180° \\ \Rightarrow x+112+x=180° \\ \Rightarrow 2x=68° \\ \Rightarrow x=34° \\ \therefore \angle PQA=34° \end{gathered}$$