In triangle $A B C$, the bisectors of its interior angles intersect at point $S$. The angle $B S C$ has a size of $130^{\circ}$, and the angle $A S C$ has $120^{\circ}$. What are the sizes of all the interior angles in triangle $A B C$?

(E. Patáková)
Let's denote the internal angles of the triangle as $\alpha, \beta$, and $\gamma$, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_64ff2634ae8ea3e7d48ag-1.jpg?height=771&width=640&top_left_y=1622&top_left_x=705)

The sum of the internal angles in any triangle is $180^{\circ}$. Therefore, in triangle $A S C$ we have

$$
\frac{\alpha}{2}+\frac{\gamma}{2}+120^{\circ}=180^{\circ}
$$

from which we can calculate

$$
\begin{aligned}
\frac{\alpha}{2}+\frac{\gamma}{2} & =60^{\circ} \\
\alpha+\gamma & =120^{\circ}
\end{aligned}
$$

In triangle $A B C$, we have $\alpha+\beta+\gamma=180^{\circ}$, from which we can now express the angle $\beta$:

$$
\beta=180^{\circ}-(\alpha+\gamma)=180^{\circ}-120^{\circ}=60^{\circ}.
$$

Similarly, in triangle $B S C$:

$$
\begin{aligned}
\frac{\gamma}{2}+\frac{\beta}{2}+130^{\circ} & =180^{\circ} \\
\frac{\gamma}{2}+\frac{\beta}{2} & =50^{\circ} \\
\gamma+\beta & =100^{\circ}
\end{aligned}
$$

Since $\beta=60^{\circ}$, it must be that $\gamma=100^{\circ}-60^{\circ}=40^{\circ}$ and finally $\alpha=120^{\circ}-40^{\circ}=80^{\circ}$ (alternatively $\alpha=180^{\circ}-100^{\circ}=80^{\circ}$).

Evaluation. 2 points for determining the sum $\alpha+\gamma=120^{\circ}$, or $\gamma+\beta=100^{\circ}$; 4 points for determining the magnitudes of all internal angles.