d
From Snell's law,  \(\frac{\mu_{\mathrm{R}}}{\mu_{\mathrm{D}}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\)  ..... \((i)\)

\(\because \,\angle {\text{i}} = {\text{A}}\)  and  \(\angle {\text{r}} = \left( {{{90}^o} - {\text{A}}} \right)\)

We also know that, \(\sin \theta_{C}=\frac{\mu_{R}}{\mu_{D}}\)

From \(eq^{n}\,\,(i)\),   \(\sin {\theta _C} = \frac{{\sin A}}{{\sin \left( {{{90}^o} - A} \right)}}\)

\(\sin {\theta _{\text{C}}} = \frac{{\sin {\text{A}}}}{{\cos {\text{A}}}}\)

\(\sin {\theta _{\text{C}}} = \tan {\text{A}}\)

\({\text{A}} = {\tan ^{ - 1}}\left( {\sin {\theta _{\text{C}}}} \right)\)