$17 \cdot 7$ In $\triangle A B C$, $\angle A=100^{\circ}, \angle B=50^{\circ}, \angle C=30^{\circ}$, and $A H$ is an altitude, $B M$ is a median, then $\angle M H C$ equals
(A) $15^{\circ}$.
(B) $22.5^{\circ}$.
(C) $30^{\circ}$.
(D) $40^{\circ}$.
(E) $45^{\circ}$.
(40th American High School Mathematics Examination, 1989)
[Solution] In the right triangle $\triangle A H C$, $M$ is the midpoint of $A C$, so $H M$ is the median to the hypotenuse, thus $M H = M C$.
$\therefore$
$$
\angle M H C = \angle C = 30^{\circ} \text{.}
$$

Therefore, the answer is $(C)$.