2. Given that $H$ is the orthocenter of $\triangle A B C$ and $B H=\sqrt{3} A C$. Then the degree measure of $\angle A B C$ is $\qquad$ .
$2.30^{\circ}$ or $150^{\circ}$.
Obviously, $\angle A B C$ cannot be a right angle. Otherwise, $B$ would be the orthocenter. Let $A C=a$. Then $B H=\sqrt{3} a$.

If $\angle A B C$ is acute, let $A H$ intersect $B C$ at point $D$, and $B H$ intersect $A C$ at point $E$. Then $A D \perp B C, B H \perp A C$.
It is easy to prove that Rt $\triangle B D H \backsim \operatorname{Rt} \triangle A D C$
$$
\begin{array}{l}
\Rightarrow \frac{B D}{A D}=\frac{B H}{A C}=\frac{\sqrt{3} a}{a}=\sqrt{3} \\
\Rightarrow \cot \angle A B C=\frac{B D}{A D}=\sqrt{3} \\
\Rightarrow \angle A B C=30^{\circ} .
\end{array}
$$

If $\angle A B C$ is obtuse, as shown in Figure 7, extend $H B$ to intersect $A C$ at point $E$, and connect $A H$ to intersect the extension of $C B$ at point $D$. It is easy to know
$$
B E \perp A C, A D \perp B C .
$$
$$
\begin{array}{c}
\text { It can be proven that Rt } \triangle B D H \backsim \\
\operatorname{Rt} \triangle A D C \\
\Rightarrow \frac{B D}{A D}=\frac{B H}{A C}=\sqrt{3} \\
\Rightarrow \cot \angle A B D=\frac{B D}{A D}=\sqrt{3} \\
\Rightarrow \angle A B D=30^{\circ} \\
\Rightarrow \angle A B C=180^{\circ}-\angle A B D=150^{\circ} .
\end{array}
$$