12.168. In a triangle, the side $a$, the angle $\alpha$ opposite to it, and the height $h$ drawn to the given side are given. Find the sum of the other two sides.
## Solution.

Let $b$ and $c$ be the unknown sides of the triangle. Then the area of the triangle $S=\frac{1}{2} b c \sin \alpha=\frac{1}{2} a h, b c=\frac{a h}{\sin \alpha}$.

$$
\begin{aligned}
& a^{2}=b^{2}+c^{2}-2 b c \cos \alpha \Leftrightarrow b^{2}+c^{2}=a^{2}+2 b c \cos \alpha \Leftrightarrow(b+c)^{2}= \\
& =b^{2}+c^{2}+2 b c=a^{2}+2 b c \cos \alpha+2 b c=a^{2}+2 b c(1+\cos \alpha)= \\
& =a^{2}+\frac{2 a h}{\sin \alpha} \cdot \cos ^{2} \frac{\alpha}{2}=a^{2}+\frac{4 a h \cos ^{2} \frac{\alpha}{2}}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=a^{2}+2 a h \operatorname{ctg} \frac{\alpha}{2} . \\
& \text { Answer: } \sqrt{a^{2}+2 a h \operatorname{ctg} \frac{\alpha}{2}} .
\end{aligned}
$$