Question

In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If $\angle$PQR$={136}^o$, then $\angle$PQA is __________.

• A. ${24}^o$
• B. ${22}^o$
• C. ${26}^o$
• D. ${21}^o$

Answer 1

The correct option is B ${22}^o$

The law of reflection states that the incident ray, the reflected ray and the normal to the surface of reflection all lie in the same plane and the angle of reflection is equal to the angle of incidence.

So, $\angle PQA=\angle PQB$

$\because$ $AB$ is a straight line

$\Rightarrow$ $\angle AQB$ =${180}^{\circ }$

$\Rightarrow$$\angle PQA+\angle PQAR+\angle PQB$ $=$${180}^{\circ }$ $(\because$ $\angle AQB$ $=$ $\angle PQA+\angle PQAR+\angle PQB)$

$\Rightarrow$ $\angle PQA$ $+$ ${136}^{\circ }$ $+$ $\angle PQA$ $=$${180}^{\circ }$

$\Rightarrow$ $2$ $\angle PQA$ $=$ ${180}^{\circ }-$ ${136}^{\circ }$

$\Rightarrow$ $\angle PQA=$ ${22}^{\circ }$

Hence the option $\left(B\right)$ ${22}^{\circ }$ is correct.