Question

In the figure given, AB is a mirror, PQ is the incident ray and QR, the reflected ray. If $\angle$PQR $={112}^o$, find $\angle$PQA.

Answer 1

$Byla{w}^{\prime }ofreflectiontheangleofincidenceisequaltoanglereflection\therefore \angle PQA+\angle PQR+\angle RQB={180}^{0}Q+{112}^0+Q={108}^{0}2Q+{112}^0={180}^{0}2Q={180}^0{112}^{0}2Q=68Q={34}^0\therefore \angle PQA={34}^0$