Problem 5. Let $\triangle A B C$ be a right triangle with a right angle at vertex $C$ and the angle at vertex $B$ is $15^{\circ}$. If $\overline{A B}=20 \mathrm{~cm}$, find the area of $\triangle A B C$.
Solution. Let $C C_{1}$ be the median in $\triangle A B C$, and $C C_{2}$ be the altitude. Then $\triangle C C_{1} B$ and $\triangle A C C_{1}$ are isosceles triangles, with base angles of $15^{\circ}$ and $75^{\circ}$, respectively. From $\measuredangle A C C_{2}=15^{\circ}$ and $\measuredangle A C C_{1}=75^{\circ}$ it follows that

$$
\measuredangle C_{2} C C_{1}=60^{\circ} \text { i.e. } \measuredangle C C_{1} C_{2}=30^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_8aa76aa53b232df96dffg-10.jpg?height=286&width=947&top_left_y=679&top_left_x=382)

Triangle $C C_{1} C_{2}$ is a right triangle and $C C_{2}$ is the leg opposite the $30^{\circ}$ angle, so it is half the hypotenuse, i.e., $\overline{C C_{2}}=\frac{\overline{C C_{1}}}{2}=5 \mathrm{~cm}$. It follows that

$$
P_{\triangle A C_{1} C}=\frac{\overline{A C_{1}} \cdot \overline{C C_{2}}}{2}=\frac{5 \cdot 10}{2}=25 \mathrm{~cm}^{2}
$$

Finally

$$
P_{\triangle A B C}=2 P_{\triangle A C_{1} C}=50 \mathrm{~cm}^{2}
$$

## IX Division