A diver at a depth of $12m$ in water $\left(\mu =\frac{4}{3}\right)$ sees the sky in a cone of semi-vertical angle of :

Given : $\mu =\frac{4}{3}$ and ${\mu }_{air}=1$

The ray emerges out of water  making an angle $90^o$ with the water surface, i.e $r=90^o$  

Using Snell's law of refraction at point A:

$\Rightarrow \mu sin\theta ={\mu }_{air}sinr$

$\Rightarrow$$\frac{4}{3}\times sin\theta =1\times sin90^o$

$\Rightarrow \frac{4}{3}\times sin\theta =1\times 1$             

$\Rightarrow$ $sin\theta =\frac{3}{4}$

$\Rightarrow$ $\theta =si{n}^{-1}\left(\frac{3}{4}\right)$