By Snell's law

\(1 \sin 60^{\circ}=\sqrt{3} \sin r\)

\(\frac{\sqrt{3}}{2}=\sqrt{3} \sin r\)

\(\sin r=\frac{1}{2}\)

\(r=30^{\circ}\)

Angle between refracted and reflected ray is \(90^{\circ}\)

Method \((ii)\)

Because angle of incidence is Brewster's angle so that angle between reflected and refracted ray is \(90^{\circ}\)

\(\operatorname{tani}_{p}=\mu=\sqrt{3}\)

\(i_{p}=60^{\circ}= i\)