2. In $\triangle A B C$, segment $A D$ intersects $B C$ at $D$, and $A D$ divides $\triangle A B C$ into two similar triangles, with a similarity ratio of $\sqrt{3}: 3$. Then the interior angles of $\triangle A B C$ are $\qquad$ .
2. $\angle B A C=90^{\circ}, \angle B=30^{\circ}, \angle C=60^{\circ}$.

As shown in Figure 3, since $\angle A D C \neq$
$$
\begin{array}{l}
\angle B, \angle A D C \neq \angle D A B, \\
\angle A D B \neq \angle C, \angle A D B \neq \\
\angle D A C \text { (these four inequalities can only be } \\
\text { greater), and } \angle A D B+\angle A D C= \\
180^{\circ}, \triangle A D B \backsim \triangle C D A, \\
\therefore \angle A D B=\angle A D C=90^{\circ} .
\end{array}
$$
(i) If $\angle B A D=\angle C A D$, then $\triangle A D B \cong \triangle A D C$.

This contradicts the similarity ratio of $\sqrt{3}: 3$.
(ii) If $\angle B A D=\angle C$, then $\angle B A C=\angle C+\angle C A D$ $=90^{\circ}, \frac{A C}{A B}=\frac{\sqrt{3}}{3}=\operatorname{tg} B$, hence $\angle C=60^{\circ}, \angle B=30^{\circ}$.