5. On the altitude $A H$ of the regular tetrahedron $A B C D$, take the midpoint $M$, and connect $B M 、 C M$, then $\angle B M C=$ $\qquad$ .
5. $90^{\circ}$.

Let the edge length of the regular tetrahedron $ABCD$ be $a$. As shown in Figure 10, connect $BH$ intersecting $CD$ at $E$, then $BE$ is the median of the equilateral $\triangle BCD$, so $BE = \frac{\sqrt{3}}{2} \cdot a$.
Furthermore, we have
$$
\begin{array}{l} 
B H = \frac{2}{3} B E \\
= \frac{\sqrt{3}}{3} a, \\
A H = \sqrt{A B^{2} - B H^{2}} = \frac{\sqrt{6}}{3} a, \\
M H = \frac{1}{2} A H = \frac{\sqrt{6}}{6} a, B M^{2} = B H^{2} + M H^{2} = \frac{1}{2} a^{2} .
\end{array}
$$

Thus, $B M^{2} + C M^{2} = 2 B M^{2} = a^{2} = B C^{2}$.
By the converse of the Pythagorean theorem, $\triangle B M H$ is a right triangle, $\angle B M C = 90^{\circ}$.