## Task B-1.5.

Let $A, B, C$ be points on a circle with center at point $S$ such that $\angle S B C=30^{\circ}$, and $\angle B C A=50^{\circ}$. What is the measure of the angle $\angle A B C$? (Point $S$ is inside the triangle $A B C$).
## First Solution.

Triangle $B S C$ is isosceles, so the angle $\angle B S C=120^{\circ}$.

In the isosceles triangle $S C A$, the angle $\angle S C A=50^{\circ}-30^{\circ}=20^{\circ}$, and the angle $\angle A S C=140^{\circ}$.

Furthermore, the angle $\angle A S B=360^{\circ}-(\angle B S C+\angle A S C)=360^{\circ}-\left(120^{\circ}+140^{\circ}\right)=100^{\circ}$. (1 point)

Finally, $\angle A B S=\frac{1}{2}\left(180^{\circ}-100^{\circ}\right)=40^{\circ}$ and $\angle A B C=40^{\circ}+30^{\circ}=70^{\circ}$.