The angle of incidence equals $60^{\circ }$ and $\tan 60^{\circ }=\sqrt{3}=\mu$, which is exactly the Brewster condition.

At Brewster's angle, the reflected and refracted rays are perpendicular to each other.

Equivalently, by Snell's law, $\sin 60^{\circ }=\sqrt{3}\sin r$ gives $\sin r=\frac{1}{2}$, so $r=30^{\circ }$. The reflected ray makes $60^{\circ }$ with the normal on the same side, and the refracted ray makes $30^{\circ }$ on the other side, so the angle between them is $180^{\circ }-60^{\circ }-30^{\circ }=90^{\circ }$.