12.54. 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 $\angle B A C=75^{\circ}$.
12.54. Let $B'$ be the point of intersection of the perpendicular bisector of segment $AC$ with line $AB$. Then $AB' = CB'$ and $\angle AB'C = 180^\circ - 2 \cdot 75^\circ = 30^\circ$. Therefore, $AB' = CB' = 2CH = AB$, i.e., $B' = B$ and $\angle B = 30^\circ$.