Let the air and glass be medium 1 and 2 , respectively. By Snell's law,
$$\begin{array}{l}n_{2} \sin \theta_{2}=n_{1} \sin \theta_{1} \\Or \quad 1.56 \sin \theta_{2}=\sin \theta_{1}\end{array}$$
But the conditions of the problem are such that $$ \theta_{1}=2 \theta_{2}, $$ so $$ 1.56 \sin \theta_{2}= $$ $$ \sin 2 \theta_{2} $$ We now use the double-angle trig identity suggested:
$$\begin{array}{l}1.56 \sin \theta_{2}=2 \sin \theta_{2} \cos \theta_{2} \\Or \quad \cos \theta_{2}=\dfrac{1.56}{2}=0.780\end{array}$$
Thus,$$\theta_{2}=38.7^{\circ} \text { and } \theta_{1}=2 \theta_{2}=77.5^{\circ}$$