Let $${\widehat v_i} = {1 \over 2}\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ and $${\widehat v_r} = {1 \over 2}\left( {\widehat i - \sqrt 3 \widehat j} \right)$$ be the unit vectors along the incident and the reflected rays. From Snell's law, angle of incidence (say $$\theta$$) is equal to the angle of reflection. Thus, angle between $$ - {\widehat v_i}$$ and $${\widehat v_r}$$ is 2$$\theta$$.

Using, dot products of vectors, we get

$$\cos (2\theta ) = - {\widehat v_i}\,.\,{\widehat v_r} = 1/2$$.

Hence, $$\theta = 30^\circ $$.