In a right-angled triangle, the exterior angles lying on the hypotenuse are in the ratio of $9: 11$. What are the angles of the triangle?
Let one of the acute angles be $\alpha$. Then one of the exterior angles lying on the hypotenuse is $180^{\circ}-\alpha$ and the other is $180^{\circ}-\left(90^{\circ}-\alpha\right)=90^{\circ}+\alpha$, thus

$$
\frac{180^{\circ}-\alpha}{90^{\circ}+\alpha}=\frac{9}{11}
$$

from which

$$
\alpha=58^{\circ} 30', \quad \beta=31^{\circ} 30'
$$

(Sándor Molnár, Zenta.)

Number of solutions: 31.