12.192. Given two sides $a$ and $b$ of a triangle and the bisector $l$ of the angle between them. Find this angle.
Solution.

Let $B D$ be the bisector of $\triangle A B C, B D=l, B C=a, A B=b$ (Fig. 12.59), $\angle A B C=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0941.jpg?height=462&width=624&top_left_y=76&top_left_x=366)

Fig. 12.59

Then $\angle A B D=\angle C B D=\frac{\alpha}{2}$.

$$
\begin{aligned}
& S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle C B D} \Rightarrow \frac{a b}{2} \sin \alpha=\frac{a l}{2} \sin \frac{\alpha}{2}+\frac{b l}{2} \sin \frac{\alpha}{2} \Leftrightarrow 2 a b \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}= \\
& =l \sin \frac{\alpha}{2} \cdot(a+b) \Leftrightarrow \cos \frac{\alpha}{2}=\frac{l(a+b)}{2 a b} ; \alpha=2 \arccos \frac{l(a+b)}{2 a b}
\end{aligned}
$$

Answer: $2 \arccos \frac{l(a+b)}{2 a b}$.