(5) Let $H$ be the orthocenter of acute triangle $A B C$, given that $\angle A=30^{\circ}, B C=3$, then $A H=$ $\qquad$ .
In $\triangle A B C$, by the Law of Sines, we have
$$
\frac{B C}{\sin \angle B A C}=\frac{A B}{\sin \angle C} \text {. }
$$

In $\triangle A B H$, by the Law of Sines, we have
$$
\frac{A B}{\sin \angle A H B}=\frac{A H}{\sin \angle A B H} \text {. }
$$

Since $H$ is the orthocenter of $\triangle A B C$, we have $\angle A H B=180^{\circ}-\angle C$, $\angle A B H=90^{\circ}-\angle B A C$, so
$$
\begin{aligned}
& \frac{B C}{\sin \angle B A C}=\frac{A B}{\sin \angle C}=\frac{A B}{\sin \angle A H B} \\
= & \frac{A H}{\sin \angle A B H}=\frac{A H}{\cos \angle B A C},
\end{aligned}
$$

Therefore, $A H=B C \cdot \cot \angle B A C=3 \cdot \cot 30^{\circ}=3 \sqrt{3}$.