In a triangle two altitudes are not smaller than the sides on to which they are dropped. Find the angles of the triangle.
1. Let's denote the triangle as \( \triangle ABC \) with sides \( a, b, \) and \( c \) opposite to vertices \( A, B, \) and \( C \) respectively. Let the altitudes from vertices \( A, B, \) and \( C \) be \( h_a, h_b, \) and \( h_c \) respectively.
2. According to the problem, two altitudes are not smaller than the sides onto which they are dropped. Without loss of generality, assume \( h_a \geq a \) and \( h_b \geq b \).
3. Recall that the altitude \( h_a \) from vertex \( A \) to side \( a \) is given by:
   \[
   h_a = b \sin C
   \]
   Similarly, the altitude \( h_b \) from vertex \( B \) to side \( b \) is given by:
   \[
   h_b = a \sin C
   \]
4. Given \( h_a \geq a \) and \( h_b \geq b \), we have:
   \[
   b \sin C \geq a \quad \text{and} \quad a \sin C \geq b
   \]
5. Dividing the first inequality by \( a \) and the second by \( b \), we get:
   \[
   \frac{b \sin C}{a} \geq 1 \quad \text{and} \quad \frac{a \sin C}{b} \geq 1
   \]
6. Simplifying these inequalities, we obtain:
   \[
   \sin C \geq \frac{a}{b} \quad \text{and} \quad \sin C \geq \frac{b}{a}
   \]
7. Since \( \sin C \) must be the same in both inequalities, we conclude:
   \[
   \sin C \geq \max\left(\frac{a}{b}, \frac{b}{a}\right)
   \]
8. The maximum value of \( \sin C \) is 1, which occurs when \( C = 90^\circ \). Therefore, for the inequalities to hold, \( \frac{a}{b} \) and \( \frac{b}{a} \) must both be 1, implying \( a = b \).
9. If \( a = b \), then \( \triangle ABC \) is an isosceles right triangle with \( C = 90^\circ \) and \( A = B = 45^\circ \).

The final answer is \( \boxed{ 45^\circ, 45^\circ, 90^\circ } \)