A right-angled triangle's exterior angles lying on the hypotenuse are in the ratio of 9:11. What are the angles of the triangle?
Let the acute angles of a right-angled triangle be denoted by $\alpha$ and $\beta$, then the exterior angles are $180^{\circ}-\alpha$, and respectively $180^{\circ}-\beta=$ $180^{\circ}-\left(90^{\circ}-\alpha\right)=90^{\circ}+\alpha$

According to the problem,

$$
\left(180^{\circ}-\alpha\right):\left(90^{\circ}+\alpha\right)=9: 11
$$

which means

$$
810^{\circ}+9 \alpha=1980^{\circ}-11 \alpha
$$

thus

$$
20 \alpha=1170^{\circ}
$$

from which

$$
\alpha=58.5^{\circ} \quad \text { and thus } \quad \beta=31.5^{\circ}
$$

Maria Toth (Miskolc, 13th grade, Mechanical Engineering Technology II. class)