\begin{aligned} & {[\text { The inscribed angle is half the central angle ] }} \\ & \text { [angles subtended by equal arcs and equal chords }]\end{aligned}

A circle is circumscribed around an equilateral triangle $ABC$. On the arc $BC$ that does not contain point $A$, there is a point $M$ that divides the degree measure of this arc in the ratio $1:2$. Find the angles of triangle $AMB$.
Let $B M: C M=1: 2$.

$$
B M=\frac{1}{3} B M C=\frac{1}{3} \cdot 120^{\circ}=40^{\circ}
$$

By the inscribed angle theorem,

$$
\angle B A M=\frac{1}{2} B M=20^{\circ},
$$

and since the inscribed angles $A M B$ and $A C B$ subtend the same arc, $\angle A M B=\angle A C B=60^{\circ}$. Therefore,

$$
\angle A B M=180^{\circ}-20^{\circ}-60^{\circ}=100^{\circ} .
$$

If $B M: C M=2: 1$, then similarly we find that

$$
\angle B A M=40^{\circ}, \angle A M B=60^{\circ} \angle A B M=80^{\circ}
$$

## Answer

$20^{\circ}, 60^{\circ}, 100^{\circ}$ or $40^{\circ}, 60^{\circ}, 80^{\circ}$.