Question

In the given figure, angle $\theta$ is

• A. ${40}^{\circ }$
• B. ${60}^{\circ }$
• C. ${120}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is D ${80}^{\circ }$

Incident ray made angle ${40}^{\circ }$ with the mirror $M2$
Angle of incidence $={90}^{\circ }-{40}^{\circ }={50}^{\circ }$
Using Law of reflection
Angle of incidence $\angle NCO=$ Angle of reflection $\angle OCA$
So angle of reflection $\angle OCA={50}^{\circ }$
$\angle ACB={90}^{\circ }-{50}^{\circ }={40}^{\circ }$
Using triangle law,
Sum of all the interior angles of triangle $={180}^{\circ }$
$\angle BAC={180}^{\circ }-({40}^{\circ }+{40}^{\circ })={100}^{\circ }$
Angle $\theta ={180}^{\circ }-{100}^{\circ }={80}^{\circ }$.