4. Let $H$ be the orthocenter of acute triangle $A B C$, given that $\angle A=60^{\circ}, B C=3$, then $A H=$
4. $\sqrt{3}$ Detailed Explanation: In $\triangle A B C$, by the Law of Sines, we have $\frac{B C}{\sin \angle B A C}=\frac{A B}{\sin \angle C}$; in $\triangle A B H$, by the Law of Sines, we have $\frac{A B}{\sin \angle A H B}=\frac{A H}{\sin \angle A B H} ; \because H$ is the orthocenter of the acute triangle $A B C$, $\therefore \angle A H B=180^{\circ}-\angle C$ and $\angle A B H=90^{\circ}-$
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\angle B A C, \therefore \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}, \therefore A H=B C \cdot \cot \angle B A C=3 \cot 60^{\circ}=\sqrt{3} .
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