3. Let $O$ be the center of the circumcircle of triangle $ABC$, points $O$ and $B$ lie on opposite sides of line $AC$, $\angle AOC = 60^\circ$. Find the angle $AMC$, where $M$ is the center of the incircle of triangle $ABC$.
# Solution.

Since points $O$ and $B$ lie on opposite sides of line $A C$, the degree measure of arc $A C$, not containing point $B$, is $360^{\circ}-60^{\circ}=300^{\circ}$. Therefore,
$\angle A B C=(1 / 2) \cdot 300^{\circ}=150^{\circ}$. The sum of the angles at vertices $A$ and $C$ of triangle $A B C$ is $180^{\circ}-150^{\circ}=30^{\circ}$, and since $A M$ and $C M$ are the angle bisectors of triangle $A B C$, the sum of the angles at vertices $A$ and $C$ of triangle $A M C$ is $15^{\circ}$. Therefore, $\angle A M C=180^{\circ}-15^{\circ}=165^{\circ}$.

Answer: $165^{\circ}$.

Instructions. Only the answer -0 points; found $\angle A B C-1$ point.