The plane mirrors $M_1$ and $M_2$ are inclined to each other such that a ray of light incident on mirror $M_1$ and parallel to the mirror $M_2$ is reflected from mirror $M_2$ parallel to the mirror $M_1$. The angle between the two mirror is (in degrees)

Given condition is shown in the figure given below, where two plane mirrors inclined to each other, such that a ray of light incident on the first mirror$\left(M_1\right)$ and parallel to the second mirror $\left(M_2\right)$ is finally reflected from second mirror $\left(M_2\right)$ parallel to the first mirror.

where, PQ = incident ray parallel to the mirror $\left(M_2\right)$,
QR = reflected ray from the mirror $\left(M_1\right)$,
RS = reflected ray from the mirror $\left(M_2\right)$ which is parallel to the $\left(M_1\right)$ and
$\theta$ = angle between $\left(M_1\right)$ and $\left(M_2\right)$.
According to geometry,
$\angle PAS=\angle PQ{M}_1=\theta$ (angle on same line)
$\angle AQ{N}_1$ = angle of incidence = 90 - $\theta$
$\angle N_{1}QR$ = angle of reflection = $(90-\theta )$
therefore, for triangle $\Delta$ORQ, (according to geometry)
$\angle \theta$ + $\angle \theta$ +$\angle ORQ={180}^o$
$\angle ORQ={180}^o-2\theta$ ...(i)
For normal $N_2$, angle of incidence = angle of reflection
$=2\theta -{90}^o$ ...(ii)
therefore, for the trianlge $\Delta$RAQ
$$\Rightarrow 4\theta -{180}^o+{180}^o-2\theta +\theta ={180}^o$$
$3\theta ={180}^o$
$\theta ={60}^o$

Why this question? Note: The key to solve such questions is to apply the reflection laws at all the reflecting surfaces along with the specific condition that the question demands.