A line through a vertex of a non-degenerate triangle cuts it in two similar triangles with $\sqrt{3}$ as the ratio between correspondent sides.
Find the angles of the given triangle.
1. **Let the line \( AD \) divide \( \triangle ABC \) into two similar triangles \( \triangle ABD \) and \( \triangle ADC \) with \( D \) on \( BC \).**

2. **Since the triangles are similar, the corresponding angles are equal. Let the angles of \( \triangle ABC \) be \( \alpha, \beta, \gamma \) such that \( \alpha + \beta + \gamma = 180^\circ \).**

3. **Given that the ratio of the corresponding sides of the similar triangles is \( \sqrt{3} \), we have:**
   \[
   \frac{BD}{DA} = \sqrt{3} \quad \text{or} \quad \frac{DA}{DC} = \sqrt{3}
   \]

4. **Consider the case where \( \frac{BD}{DA} = \sqrt{3} \). This implies that \( \triangle ABD \) and \( \triangle ADC \) are similar with a ratio of \( \sqrt{3} \).**

5. **Since \( \triangle ABD \sim \triangle ADC \), the corresponding angles are equal:**
   \[
   \angle ABD = \angle ACD \quad \text{and} \quad \angle BAD = \angle CAD
   \]

6. **If \( \angle ABD = \angle ACD \), then \( \angle ADB = 90^\circ \) because the sum of angles in a triangle is \( 180^\circ \).**

7. **Now, consider the angles in \( \triangle ABC \). Since \( \angle ADB = 90^\circ \), \( \angle ABD \) and \( \angle ACD \) must be complementary to \( 90^\circ \).**

8. **Let \( \angle ABD = \theta \). Then \( \angle ACD = \theta \) and \( \angle BAD = 90^\circ - \theta \).**

9. **Since \( \triangle ABC \) is divided into two similar triangles, the angles must satisfy the condition:**
   \[
   \alpha = 90^\circ, \quad \beta = 30^\circ, \quad \gamma = 60^\circ
   \]

10. **Thus, the angles of \( \triangle ABC \) are \( 30^\circ, 60^\circ, \) and \( 90^\circ \).**

\(\blacksquare\)

The final answer is \( \boxed{ 30^\circ, 60^\circ, 90^\circ } \)