[ Auxiliary circle ] [ Law of Cosines ]

In triangle $ABC$, a point $M$ is taken on side $BC$, such that $BM = 2MC$ and $\angle AMB = 60^{\circ}$. Given that $\angle BAC = 60^{\circ}$, find the angles $B$ and $C$ of triangle $ABC$.
The foot of the perpendicular dropped from point $B$ to $A M$ is the center of the circle circumscribed around triangle $A B C$.

## Solution

Let point $O$ be the foot of the perpendicular dropped from point $B$ to $A M$. Then

$$
O M=\frac{1}{2} B M=M C, \angle M O C=\angle O C M=30^{\circ}, \angle B O C=120^{\circ}
$$

Therefore, point $O$ is the center of the circle circumscribed around triangle $A B C$. Hence,

$$
\angle A C B=\frac{1}{2} \angle A O B=45^{\circ} .
$$

## Answer

$75^{\circ} ; 45^{\circ}$.

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