8. The length of the hypotenuse of a right-angled triangle is $15 \mathrm{~cm}$, and one of its angles is $15^{\circ}$. Calculate (20) the length of the segment of the angle bisector of the right angle that lies within the triangle.
First solution. The bisector of the right angle intersects the hypotenuse at point $D$. Let $s$ be the length of the segment of the bisector of the right angle within the triangle.

Consider triangles $A B C$ and $A D C$:

![](https://cdn.mathpix.com/cropped/2024_05_30_fac652c8a845a57c6fa7g-17.jpg?height=312&width=534&top_left_y=221&top_left_x=772)

\[
\begin{gathered}
\cos 15^{\circ}=\frac{b}{c} \quad \text { i.e. } \quad b=c \cos 15^{\circ}=15 \cos 15^{\circ} \\
\frac{b}{s}=\frac{\sin \left(180^{\circ}-\left(45^{\circ}+15^{\circ}\right)\right)}{\sin 15^{\circ}}=\frac{\sin 120^{\circ}}{\sin 15^{\circ}}=\frac{\sin 60^{\circ}}{\sin 15^{\circ}}=\frac{\sqrt{3}}{2 \sin 15^{\circ}}
\end{gathered}
\]

(10 points)

from which we get

\[
s=\frac{2 b \sin 15^{\circ}}{\sqrt{3}}=\frac{2 \cdot 15 \cdot \cos 15^{\circ} \sin 15^{\circ}}{\sqrt{3}}=\frac{15 \cdot \sin 30^{\circ}}{\sqrt{3}}=\frac{15}{2 \sqrt{3}}=\frac{5 \sqrt{3}}{2}
\]

(8 points)

The length of the segment of the bisector of the right angle is $s=\frac{5 \sqrt{3}}{2} \mathrm{~cm}$.

(2 points)