[Relationships between the sides and angles of triangles (other).]

Find the angle $B$ of triangle $A B C$, if the length of the height $C H$ is half the length of side $A B$, and $\angle B A C=75^{\circ}$.
Let $B^{\prime}$ be the point of intersection of the perpendicular bisector of segment $A C$ with line $A B$. Then $A B^{\prime}=C B^{\prime}$ and $\angle$ $A B^{\prime} C=180^{\circ}-2 \cdot 75^{\circ}=30^{\circ}$. Therefore, $A B^{\prime}=C B^{\prime}=2 C H=A B$, i.e., $B^{\prime}=B$ and $\angle B=30^{\circ}$.