3. Given in $\triangle A B C$, $A C \geqslant A B$, side $B C$ is divided into $n$ ($n$ is an odd number) equal parts. Let $\alpha$ represent the angle subtended at point $A$ by the segment containing the midpoint of side $B C$, $h$ be the altitude from $A$ to side $B C$, and $B C=a$. If $\tan \alpha=\frac{4 n h}{\left(n^{2}-1\right) a}$, then $\angle B A C=$ $\qquad$
3. $90^{\circ}$.

Draw a perpendicular from point $A$ to $BC$, with the foot of the perpendicular being $D$. Let the segment containing the midpoint of $BC$ be $B' C'$. Then
$$
B B'=\frac{n-1}{2 n} a, B C'=\frac{n+1}{2 n} a .
$$

Let $\overrightarrow{B D}=x \overrightarrow{B C}(x \in \mathbf{R})$. Then
$$
D B'=\left(\frac{n-1}{2 n}-x\right) a, D C'=\left(\frac{n+1}{2 n}-x\right) a \text {. }
$$

Thus, $\tan \angle C' A D=\frac{\left(\frac{n+1}{2 n}-x\right) a}{h}$,
$$
\tan \angle B' A D=\frac{\left(\frac{n-1}{2 n}-x\right) a}{h} \text {. }
$$

Therefore, $\tan \alpha=\tan \angle C' A B'$
$$
\begin{array}{l}
=\frac{\left[\left(\frac{n+1}{2 n}-x\right)-\left(\frac{n-1}{2 n}-x\right)\right] a h}{h^{2}+\left(\frac{n+1}{2 n}-x\right)\left(\frac{n-1}{2 n}-x\right) a^{2}} \\
=\frac{4 n h a}{\left(n^{2}-1\right) a^{2}+4 n^{2}\left[h^{2}-x(1-x) a^{2}\right]} .
\end{array}
$$

Given $\tan \alpha=\frac{4 n h}{\left(n^{2}-1\right) a}$, we know
$$
h^{2}=x a \cdot(1-x) a \text {. }
$$

Thus, $A D^{2}=B D \cdot C D$, and $D$ lies on the segment $BC$.
By the inverse of the projection theorem, $\angle B A C=90^{\circ}$.