Two plane mirrors are inclined at an angle of $$60^\circ$$ with each other. A ray of light is incident on one of the mirrors at an arbitrary angle of incidence $$\angle i$$. The ray is reflected from this mirror and falls on the second mirror where it further gets reflected parallel to the first mirror. The angle of incidence $$\angle i$$ is

Let the ray leave the first mirror making angle $$\theta$$ with it. In the triangle formed by the two mirrors and this ray, the ray meets the second mirror at $$180^\circ - 60^\circ - \theta$$, and it reflects at the same angle. For the final ray to be parallel to the first mirror, this angle must equal the $$60^\circ$$ between the mirrors (alternate angles), so $$120^\circ - \theta = 60^\circ$$ and $$\theta = 60^\circ$$. The angle of incidence on the first mirror is $$i = 90^\circ - 60^\circ = 30^\circ$$.