According to the theorem about the peripheral angle and the central angle, we have: $\angle B C A=120^{\circ}: 2=60^{\circ}$.

Let $D$ be the intersection of the line through $A$ and $M$ with the chord $B C$. Then, due to the exterior angle theorem for triangle $A D C$:

$$
\angle A D B=20^{\circ}+60^{\circ}=80^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0743.jpg?height=423&width=445&top_left_y=228&top_left_x=817)

Furthermore, $\angle B M D$ is the supplementary angle to $\angle A M B$, and thus we have:

$$
\angle B M D=180^{\circ}-120^{\circ}=60^{\circ}
$$

From (1) and (2), it follows from the theorem about the sum of the interior angles in triangle $D M B$:

$$
\angle D B M(=\angle C B M)=180^{\circ}-\left(60^{\circ}+80^{\circ}\right)=40^{\circ}
$$

Therefore, the desired angle $\angle C B M=40^{\circ}$.
## Task 3 - 230823

Let $k$ be a circle with center $M$. Three points $A, B$, and $C$ on $k$ are positioned such that the point $M$ lies inside the triangle $A B C$. Furthermore, $\overline{\angle C A M}=20^{\circ}$ and $\overline{\angle A M B}=120^{\circ}$.

Determine the size of the angle $\angle C B M$ from these conditions!