(i) Since at point N, the angle of reflection is $90°$, then $\angle {\mathit{r}}_2$ is the critica angle for the glass - air pair of media.
$\mathit{s}\mathit{i}\mathit{n}\angle {\mathit{r}}_2=\frac{1}{\mathit{\mu }}=\frac{1}{\sqrt{2}}$
$\angle \mathit{r}2=\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{1}{\sqrt{2}}\right)=45°$
(ii) $\mathit{\mu }=\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{A}+\frac{{\mathit{\delta }}_{\mathit{m}}}{2}\right)}{\mathit{s}\mathit{i}\mathit{n}\frac{\mathit{A}}{2}}$
Or, $\sqrt{2}=\frac{\mathit{s}\mathit{i}\mathit{n}\left(60°+\frac{{\mathit{\delta }}_{\mathit{m}}}{2}\right)}{\mathit{s}\mathit{i}\mathit{n}\frac{60°}{2}}$
Or, $\sqrt{2}=\frac{\mathit{s}\mathit{i}\mathit{n}\left(30°+\frac{{\mathit{\delta }}_{\mathit{m}}}{2}\right)}{\frac{1}{2}}$
Or, $0.7=\mathit{s}\mathit{i}\mathit{n}\left(30°+\frac{{\mathit{\delta }}_{\mathit{m}}}{2}\right)$
Or, $\mathit{s}\mathit{i}{\mathit{n}}^{-1}0.7=30°+\frac{{\mathit{\delta }}_{\mathit{m}}}{2}$
Or, $44.4°=30°+\frac{{\mathit{\delta }}_{\mathit{m}}}{2}$
$\therefore {\mathit{\delta }}_{\mathit{m}}=28.8°$