(d) By using $\mu = \tan\theta_{p} \Rightarrow \mu = \tan 60 = \sqrt{3}$ Also $C = \sin^{- 1}{\left( \frac{1}{\mu} \right) \Rightarrow C = \sin^{- 1}\left( \frac{1}{\sqrt{3}} \right)}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.