30.8. In $\triangle A B C, \angle A=15^{\circ}$ and $B C=4$. What is the radius of the circle circumscribing $\triangle A B C$ ?
30.8. $4 \sqrt{2+\sqrt{3}}$ or $2(\sqrt{6}+\sqrt{2})$
We use the Extended Law of Sines. Let $R$ be radius of the circle circumscribing $\triangle A B C$.
$$
2 R=\frac{4}{\sin 15^{\circ}} \quad \text { or } \quad R=\frac{2}{\sin 15^{\circ}}
$$

Depending on the formula used (either half-angle formula or difference of two angles), we have
$$
\sin 15^{\circ}=\frac{\sqrt{2-\sqrt{3}}}{2} \text { or } \quad \sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4} .
$$

Then we solve for $R$ using either of these values.