3. In $\triangle A B C$,
$$
\angle A=100^{\circ} ; \angle B=50^{\circ} \text {. }
$$
$A H$ is the altitude from $A$ to side $B C$, and $B M$ is the median from $B$ to side $A C$.
Then $\angle M H C=$
$\begin{array}{l}\text { 3. } \angle M H C=30^{\circ} \text {. } \\ \left.\text { Solution } \begin{array}{l}M \text { is the midpoint of } A C \text {, } \\ \angle A H C=90^{\circ}\end{array}\right\} \Rightarrow \\ A M=M C=M H \text {. } \\ \Rightarrow \angle M A H=\angle A H M=60^{\circ} \text {. } \\ \text { And } \angle C=180^{\circ}-150^{\circ}=30^{\circ} \text {, then } \\ \angle M H C=30^{\circ} \text {. } \\\end{array}$