4. The bisector of the interior angle at the base $\overline{B C}$ of isosceles triangle $A B C$ and the bisector of the angle between the legs intersect and form an angle of $125^{\circ} 30^{\prime}$. What are the interior angles of this triangle?
4. 

![](https://cdn.mathpix.com/cropped/2024_05_30_f844036739139e74d8f5g-1.jpg?height=311&width=266&top_left_y=2243&top_left_x=952)

Let $S$ be the intersection of the angle bisectors of $\angle ABC$ and $\angle CAB$. Since $\angle BSA = 125^\circ 30'$, it follows that

$\angle BSD = 180^\circ - 125^\circ 30' = 54^\circ 30'$.

Furthermore, from the right triangle $BDS$, it follows that $\angle SBD = 90^\circ - \angle BSD = 90^\circ - 54^\circ 30' = 35^\circ 30'$.

Since line $BS$ is the angle bisector of $\angle ABC$, we conclude that $\angle ABC = \angle BCA = 2 \cdot 35^\circ 30' = 71^\circ$.