Given angle of incidence $i = 60^{\circ}$ and refractive index $\mu = \sqrt{3}$. Since $\tan(60^{\circ}) = \sqrt{3}$, the angle of incidence is equal to the Brewster's angle. At Brewster's angle, the reflected and refracted rays are perpendicular to each other, so the angle between them is $90^{\circ}$.