12.185. In a triangle, the area $S$, side $a$, and the angle $\alpha$ opposite to it are known. Find the sum of the other two sides.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0936.jpg?height=587&width=585&top_left_y=78&top_left_x=81)

Fig. 12.53

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0936.jpg?height=446&width=472&top_left_y=229&top_left_x=741)

Fig. 12.54

Let \( x \) and \( y \) be the unknown sides of the given triangle.

\[
\begin{aligned}
& \text { Then }\left\{\begin{array}{l}
S=\frac{1}{2} x y \sin \alpha ; \\
a^{2}=x^{2}+y^{2}-2 x y \cos \alpha
\end{array} \Rightarrow x^{2}+y^{2}=a^{2}+2 x y \cos \alpha \Leftrightarrow\right. \\
& (x+y)^{2}=a^{2}+2 x y \cos \alpha+2 x y=a^{2}+2 x y(1+\cos \alpha)= \\
& =a^{2}+2 \cdot \frac{2 S}{\sin \alpha} \cdot 2 \cos ^{2} \frac{\alpha}{2}=a^{2}+4 S \operatorname{ctg} \frac{\alpha}{2} \Rightarrow x+y=\sqrt{a^{2}+4 S \operatorname{ctg} \frac{\alpha}{2}}
\end{aligned}
\]

Answer: \(\sqrt{a^{2}+4 S \operatorname{ctg} \frac{\alpha}{2}}\).