[ Angles between bisectors ]

Point $O$ is the center of the circle inscribed in triangle $A B C$. It is known that $B C=a, A C=b, \angle A O B=120^{\circ}$. Find the side $A B$.
$\angle A O B=90^{\circ}+\frac{1}{2} \angle C$.

## Solution

Since $\angle A O B=90^{\circ}+\frac{1}{2} \angle C$, then

$$
\angle C=2 \angle A O B-180^{\circ}=240^{\circ}-180^{\circ}=60^{\circ} .
$$

By the cosine theorem

$$
A B^{2}=B C^{2}+A C^{2}-2 B C \cdot A C \cos \angle C=a^{2}+b^{2}-2 a b \cdot \frac{1}{2}=a^{2}+b^{2}-a b
$$

Therefore,

$$
A B=\sqrt{a^{2}+b^{2}-a b}
$$

## Answer

$\sqrt{a^{2}+b^{2}-a b}$