Answer: 92.

Solution. Let the halves of the angles $A, B, C$ of triangle $ABC$ be denoted by $x, y$, and $z$ respectively. Then, $\angle COH = 90^{\circ} - z$ and $46^{\circ} = 90^{\circ} - z - y$. Since $x + y + z = 90^{\circ}$, we have $x = 44$ and $\angle BAC = 92^{\circ}$.
3. Variant 1.

The height $A H$ and the bisector $C L$ of triangle $A B C$ intersect at point $O$. Find the angle $B A C$, if it is known that the difference between the angle $C O H$ and half the angle $A B C$ is $46^{\circ}$.