Given a right-angled triangle. Denoting its hypotenuse by $c$, its area is $t=\frac{c^{2}}{8}$. Determine the exact values of the angles of the triangle.
I. solution. Let the height corresponding to the hypotenuse be $m$. The area of the triangle can be written in two ways: $t=\frac{c^{2}}{8}=\frac{\mathrm{cm}}{2}$, from which $c=4 m$.

The height $m$ divides the side $c$ into segments of length $x$ and $4 m-x$. Applying the altitude theorem: $m^{2}=x(4 m-x)$, hence $x^{2}-4 m x+m^{2}=0$. Solving this for $x$ gives

$$
x_{1,2}=\frac{4 m \pm \sqrt{16 m^{2}-4 m^{2}}}{2}=(2 \pm \sqrt{3}) m
$$

If the angle between the segment of length $(2+\sqrt{3}) m$ and the adjacent leg is denoted by $\alpha$, then $\operatorname{tg} \alpha=\frac{1}{2+\sqrt{3}}$. A calculator gives that in this case $\alpha=15^{\circ}$.

Indeed, using the addition theorems,

$$
\operatorname{tg} 15^{\circ}=\frac{\sin 30^{\circ}}{1+\cos 30^{\circ}}=\frac{1 / 2}{1+\sqrt{3} / 2}=\frac{1}{2+\sqrt{3}}
$$

Since $2+\sqrt{3}$ and $2-\sqrt{3}$ are reciprocals, the other angle is indeed $75^{\circ}$, so the problem has a unique solution.

II. solution. Denoting the height corresponding to the hypotenuse by $m$, the area can be written in two ways: $t=\frac{c^{2}}{8}=\frac{\mathrm{cm}}{2}$, from which $m=\frac{c}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_02806bec31732c91c65ag-1.jpg?height=241&width=569&top_left_y=1045&top_left_x=775)

Let the midpoint of segment $A B$ be $O$. By the converse of Thales' theorem, $O A=O B=O C=\frac{c}{2}$. In triangle $C O T$:

$$
\sin C O T \varangle=\frac{\frac{c}{4}}{\frac{c}{2}}=\frac{1}{2},
$$

and since the angle is acute, $C O T \varangle=30^{\circ}$. Since this angle is the exterior angle of triangle $O A C$, $C O T \varangle=O A C \varangle+O C A \varangle=30^{\circ}$. Triangle $O A C$ is isosceles, so from this $B A C \varangle=O A C \varangle=O C A \varangle=30^{\circ} / 2=15^{\circ}$ follows. The size of the third angle, $A B C \varangle$, is $75^{\circ}$.

III. solution. Denote the sides and angles of the triangle in the usual way. We know that $a=c \sin \alpha$ and $b=c \cos \alpha$. Using this,

$$
\frac{c^{2}}{8}=t=\frac{a b}{2}=\frac{c \sin \alpha \cdot c \cos \alpha}{2}
$$

from which, after dividing by $c^{2} \neq 0$ and multiplying by 4, we get

$$
\frac{1}{2}=2 \sin \alpha \cos \alpha=\sin 2 \alpha
$$

from which $2 \alpha=30^{\circ}$ or $2 \alpha=150^{\circ}$, and thus $\alpha=15^{\circ}$ or $\alpha=75^{\circ}$ follows.

Therefore, the exact values of the angles of the triangle are $90^{\circ}, 75^{\circ}$, and $15^{\circ}$.