10,11 The base of the pyramid is a right-angled triangle with an acute angle of $\frac{\pi}{8}$. Each lateral edge is equal to $\sqrt{6}$ and is inclined to the base plane at an angle of $\frac{5 \pi}{13}$. Find the volume of the pyramid.
Let $D H$ be the height of the triangular pyramid $A B C D$, where the base $A B C$ is a right triangle $A B C$ with angles $\angle A C B=\frac{\pi}{2}, \angle B A C=\frac{\pi}{3}$. Since the lateral edges of the pyramid are equal, point $H$ is the center of the circle circumscribed around the right triangle $A B C$, i.e., the midpoint of the hypotenuse $A B$. From the right triangles $A D H$ and $A B C$, we find that

\[
\begin{gathered}
D H=A D \sin \angle D A H=\sqrt{6} \sin \frac{5 \pi}{13}, A B=2 A H=2 \sqrt{6} \cos \frac{5 \pi}{13} \\
B C=A B \sin \angle B A C=A B \sin \frac{\pi}{8}, A C=A B \cos \angle B A C=A B \cos \frac{\pi}{8}
\end{gathered}
\]

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_1f91526468b1b4d32730g-27.jpg?height=92&width=1278&top_left_y=1506&top_left_x=390)

\[
\begin{aligned}
& =\frac{1}{12} \sin \frac{\pi}{4} A B 2 \sqrt{6} \sin \frac{5 \pi}{13}=\frac{\sqrt{2}}{24} \cdot 24 \cos 2 \frac{5 \pi}{13} \cdot \sqrt{6} \sin \frac{5 \pi}{13}= \\
& =\sqrt{3} \cdot 2 \cos \frac{5 \pi}{13} \sin \frac{5 \pi}{13} \cos \frac{5 \pi}{13}=\sqrt{3} \sin \frac{10 \pi}{13} \cos \frac{5 \pi}{13}
\end{aligned}
\]

## Answer

$\sqrt{3} \sin \frac{10 \pi}{13} \cos \frac{5 \pi}{13}$.