Solution. Let $\alpha$ be the angle at the base and $\beta$ be the angle at the vertex. Then from the condition of the problem, we have $\frac{\alpha}{2}+\frac{\beta}{2}+130^{\circ}=180^{\circ}$, so $\alpha+\beta=100^{\circ}$. On the other hand, for the angles in $\triangle A B C$, we have $2 \alpha+\beta=180^{\circ}$. Therefore, $\alpha=80^{\circ}$, so $\beta=20^{\circ}$.

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Problem 4. In the isosceles triangle ABC, the angle bisectors of the angle at the vertex and the angle at the base are drawn. They intersect at an angle of $130^{\circ}$. Determine the angles in the triangle.