4. Inside triangle $ABC$, where $\angle C=70^{\circ}, \angle B=80^{\circ}$, a point $M$ is taken such that triangle $CMB$ is equilateral. Find the angles $MAB$ and $MAC$.
4. Answer. $\angle M A B=20^{\circ}, \angle M A C=10^{\circ}$.

Consider the circle with center at point $M$ and radius $R=M B=M C$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_71141c75ee9ce107dde7g-2.jpg?height=335&width=302&top_left_y=1977&top_left_x=223)

$\angle A=180^{\circ}-(\angle B+\angle C)=30^{\circ}, \angle B M C=60^{\circ}$, therefore, point $A$ lies on this circle, which means the circle is the circumcircle of triangle $A B C$. Thus, $M A=M B=M C$, so $\angle M A B=\angle M B A=20^{\circ}$, $\angle M A C=\angle M C A=10^{\circ}$.