Question

Question 3
In the given figure, PQ is a mirror, AB is the incident ray and BC is the reflected ray. If $\angle ABC={46}^{\circ },$ then $\angle ABP$ is equal to

(a) ${44}^{\circ }$
(b) ${67}^{\circ }$
(c) ${13}^{\circ }$
(d) ${62}^{\circ }$

Answer 1

We know that the angle of incidence is always equal to the angle of reflection.
$\angle ABP=\angle CBQ$
i.e. a = b

Now sum of all the angles on a straight line is ${180}^{\circ }$.
$\therefore a+{46}^{\circ }+b={180}^{\circ }$
$\Rightarrow 2a={180}^{\circ }-{46}^{\circ }\Rightarrow 2a={134}^{\circ }\Rightarrow a=\frac{{134}^{\circ }}{2}={67}^{\circ }\therefore \angle ABP={67}^{\circ }$