One interior angle in a triangle measures $50^{\circ}$. What is the angle between the bisectors of the remaining two interior angles?

(L. Hozová)

Hint. You don't need to know the sizes of the remaining interior angles to solve the problem.
Consider a triangle $ABC$ with an angle of $50^{\circ}$ at vertex $A$; the unknown angles at vertices $B$ and $C$ are denoted by $\beta$ and $\gamma$. The intersection of the angle bisectors is denoted by $O$, the angle $BOC$ is denoted by $\omega$, and the angle adjacent to it is denoted by $\psi$.

![](https://cdn.mathpix.com/cropped/2024_04_17_4274193d47a1b7e4f9cbg-5.jpg?height=685&width=646&top_left_y=794&top_left_x=705)

The sum of the interior angles in any triangle is $180^{\circ}$. Therefore, in triangles $ABC$ and $OBC$ we have

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

From the second equation and the fact that $\omega$ and $\psi$ are adjacent angles, it follows that

$$
\psi=\frac{\beta}{2}+\frac{\gamma}{2}
$$

From the first equation, we can express

$$
\frac{\beta}{2}+\frac{\gamma}{2}=\frac{130^{\circ}}{2}=65^{\circ},
$$

thus the angle between the bisectors of the remaining two interior angles is $65^{\circ}$.

Note. The answer that the bisectors form an angle $\omega=180^{\circ}-65^{\circ}=115^{\circ}$ is also considered correct.