8,9

The area of triangle $ABC$ is $S, \angle BAC = \alpha, AC = b$. Find $BC$.
Apply the Law of Cosines.

## Solution

Since $S=\frac{1}{2} A B \cdot A C \sin \alpha$, then $A B=\frac{2 S}{b \sin \alpha}$. By the Law of Cosines,

$$
\begin{gathered}
B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \alpha= \\
=\frac{4 S^{2}}{b^{2} \sin ^{2} \alpha}+b^{2}-2 b \cos \alpha \cdot \frac{2 S}{b \sin \alpha}=\frac{4 S^{2}}{b^{2} \sin ^{2} \alpha}+b^{2}-4 S \cot \alpha
\end{gathered}
$$

## Answer

$\sqrt{\frac{4 S^{2}}{b^{2} \sin ^{2} \alpha}+b^{2}-4 S \cot \alpha}$