## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7c0a9615fb5cd46ea1c5g-01.jpg?height=582&width=718&top_left_y=1872&top_left_x=693)

Let the intersection of the angle bisector and line $A C$ be $V$. Calculate the angle $\gamma$ in triangle $\triangle A B C$.

$$
\gamma=180^{\circ}-\left(38^{\circ}+52^{\circ}\right)=90^{\circ}
$$

The exterior angle of angle $\beta$ is $38^{\circ}+90^{\circ}=128^{\circ}$. Then $\angle C B V=64^{\circ}$.

From the right triangle $B V C$ it follows that

$$
\angle V=90^{\circ}-64^{\circ}=26^{\circ}
$$

Note: The angle $\angle V$ can also be calculated from triangle $A B V$:

$$
\angle V=180^{\circ}-\left(38^{\circ}+52^{\circ}+64^{\circ}\right)=26^{\circ}
$$
Task B-1.2. In triangle $\triangle A B C$, the measures of the angles at vertex $A$ and vertex $B$ are $\alpha=38^{\circ}$ and $\beta=52^{\circ}$, respectively. Calculate the measure of the angle formed by the external angle bisector at vertex $B$ and the line on which side $\overline{A C}$ lies.