Task 1. Calculate the area of a right triangle if the length of its hypotenuse is $c$, and one of its acute angles is $15^{\circ}$.
Solution. Let $ABC$ be a right triangle with hypotenuse $\overline{AB}=c$ and acute angle $\measuredangle ABC=15^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_a73f0c01d543f9cc2089g-07.jpg?height=202&width=774&top_left_y=1771&top_left_x=709)
The midpoint of the hypotenuse $AB$ will be denoted by $D$. According to this, $\overline{BD}=\overline{AD}=\overline{CD}=\frac{c}{2}$, and triangles $CDB$ and $ADC$ are isosceles with bases $BC$ and $AC$. The angle $\measuredangle ADC$ is an exterior angle for triangle $CDB$, so

$$
\measuredangle ADC=2 \measuredangle DBC=2 \cdot 15^{\circ}=30^{\circ}.
$$

The segment $AE$ is the altitude of triangle $ADC$ (see the diagram). According to this, triangle $DEA$ is a right triangle with acute angles $\measuredangle ADE=30^{\circ}$ and $\measuredangle DAE=60^{\circ}$, from which we get $\overline{AE}=\frac{1}{2} \overline{AD}=\frac{1}{2} \cdot \frac{c}{2}=\frac{c}{4}$.

For triangle $CAD$ we have

$$
P_{\triangle ADC}=\frac{1}{2} \cdot \frac{c}{2} \cdot \frac{c}{4}=\frac{c^{2}}{16}
$$

Triangles $\triangle CAD$ and $\triangle CDB$ have equal areas, so

$$
P_{\triangle ABC}=P_{\triangle ADC}+P_{\triangle CDB}=2 P_{\triangle ADC}=2 \cdot \frac{c^{2}}{16}=\frac{c^{2}}{8}
$$