3[ Equilateral (Equiangular) Triangle ]

The angle with vertex $C$ is $120^{\circ}$. A circle of radius $R$ touches the sides of the angle at points $A$ and $B$. Find $A B$.
$\angle A O B=60^{\circ}$.

## Solution

Let $O$ be the center of the circle. From the equality of right triangles $A O C$ and $B O C$ (by leg and hypotenuse), it follows that

$$
\angle A C O=\angle B C O=60^{\circ},
$$

thus,

$$
\angle A O C=\angle B O C=30^{\circ}, \angle A O B=60^{\circ},
$$

therefore, triangle $A O B$ is equilateral. Consequently, $A B=A O=R$.

## Answer

## Problem