A right-angled triangle's inscribed circle's radius is 0.45 times the height corresponding to the hypotenuse. What are the acute angles of the triangle?
Solution. The height corresponding to the hypotenuse of the right-angled triangle $ABC$ is $m$, the radius of the inscribed circle is $r$, and the lengths of the tangent segments from $A$ and $B$ to the inscribed circle are $x$ and $y$ respectively. We know that $r=0.45 m=\frac{9}{20} m$.

![](https://cdn.mathpix.com/cropped/2024_05_02_848ba5baa3039b2819e3g-1.jpg?height=235&width=596&top_left_y=300&top_left_x=754)

Let's write the double area of the triangle in two ways:

$$
2 T=(x+y) m
$$

and

$$
2 T=(x+y) r+(y+r) r+(x+r) r=2(x+y+r) r
$$

From the equality of the two areas, using the relationship $r=\frac{9}{20} m$, we get

$$
(x+y) m=2\left(x+y+\frac{9}{20} m\right) \frac{9}{20} m
$$

Divide by $m \neq 0$ and perform the operations:

$$
x+y=2 \cdot \frac{9}{20}(x+y)+2 \cdot \frac{9}{20} \cdot \frac{9}{20} m
$$

From this,

$$
\frac{m}{x+y}=\frac{20}{81} ; \quad \text { i.e. } \quad \frac{m}{c}=\frac{20}{81}
$$

and substituting $m=\frac{20}{9} r$ we get $r=\frac{1}{9} c$, which means the radius of the inscribed circle is one-ninth of the hypotenuse.

Next, let's apply the Pythagorean theorem to the triangle:

$$
(x+y)^{2}=(x+r)^{2}+(y+r)^{2}
$$

Rearranging the equation, we get

$$
x y=(x+y) r+r^{2}
$$

We now know the ratio of the hypotenuse to the radius of the inscribed circle. This ratio, by similarity, determines the right-angled triangle up to similarity. The similarity, in turn, uniquely determines the angles of the triangle.

Let's choose the hypotenuse to be 9 units, then $x+y=9$, so $y=9-x$, and the radius of the inscribed circle is 1. Substituting into equation (1): $x(9-x)=10$, which simplifies to $x^{2}-9 x+10=0$,

$$
x=\frac{9+\sqrt{41}}{2}, \quad \text { and } \quad \sin \beta=\frac{x+r}{x+y}=\frac{\frac{9+\sqrt{41}}{2}+1}{9}
$$

Performing the indicated operations: $\sin \beta \approx 0.9668$ and $\beta=75.195^{\circ}$, the other acute angle of the triangle is $\alpha=14.805^{\circ}$.