Example 6 In $\triangle A B C$, $\angle A=50^{\circ}, H$ is the orthocenter of $\triangle A B C$, and $H$ does not coincide with $B$ or $C$. Then $\angle B H C=$ $\qquad$
(1994, Oriental Airlines Cup - Shanghai Junior High School Mathematics Competition)
Solution: (1) When $\triangle A B C$ is an acute triangle, as shown in the figure, $A D, B E$, and $C F$ are the three altitudes, then
$$
\begin{array}{c}
\angle E B C=90^{\circ}-\angle A C B, \\
\angle F C B=90^{\circ}-\angle A B C . \\
\text { Therefore } \begin{aligned}
\angle B H C & =180^{\circ}-\angle E B C-\angle F C B \\
= & \angle A C B+\angle A B C \\
= & 180^{\circ}-\angle A=130^{\circ} .
\end{aligned}
\end{array}
$$
(2) When $\triangle A B C$ is an obtuse triangle, assume $\angle B$ $>90^{\circ}$. As shown in the figure, we know
$$
\begin{array}{r}
\angle A B E=\angle H B D . \\
\text { Also } \angle A E B=90^{\circ}, \\
\text { Therefore } \angle B H C=\angle A=50^{\circ} .
\end{array}
$$

In summary: Therefore, the answers are $130^{\circ}$ and $50^{\circ}$.