# The answer should be ?

$\sin \frac{{90}^{\circ} - p}{\sin} \left({90}^{\circ} - q\right) = \cos \frac{p}{\cos} q$
The angles of incidence and refraction that appear in Snell's law are measured from the normal. It is easy to see that they are ${90}^{\circ} - p$ and ${90}^{\setminus} \circ - q$ , respectively.
$\sin \frac{{90}^{\circ} - p}{\sin} \left({90}^{\circ} - q\right)$