In a triangle, the side $a$ and two adjacent angles $\beta$ and $\gamma$ are known. Find the bisector drawn from the vertex of the third angle.
Apply the Law of Sines.

## Solution

Let $A M$ be the bisector of triangle $A B C, C B=a, \angle B=\beta, \angle C=\gamma$. Then $\frac{A B}{\sin \gamma}=\frac{a}{\sin (\beta+\gamma)} \Rightarrow A B=$ $\frac{a \sin \gamma}{\sin (\beta+\gamma)}$

In triangle $A M B, \angle A M B=\gamma+1 / 2 \angle A=\gamma+90^{\circ}-1 / 2(\gamma+\beta)=90^{\circ}+(\gamma-\beta), \frac{A M}{\sin \beta}=\frac{A B}{\sin \angle A M B}=\frac{\frac{a \sin \gamma}{\sin (\beta+\gamma)}}{\cos \frac{\gamma-\beta}{2}}$

From this, we find that $A M=\frac{a \sin \beta \sin \gamma}{\sin (\gamma+\beta) \cos \frac{\beta-\gamma}{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-13.jpg?height=343&width=606&top_left_y=203&top_left_x=726)

## Answer

$a \sin \beta \sin \gamma$ $\overline{\sin (\gamma+\beta) \cos \frac{\beta-\gamma}{2}}$.