10,11

The base of the pyramid is a right-angled triangle with an area of $S$. The lateral edges of the pyramid are equal to each other. The dihedral angles at the legs of its base are $\alpha$ and $\beta$. Find the volume of the pyramid.
Let the base of the tetrahedron $ABCD$ be a right triangle $ABC$ with legs $BC = a$ and $AC = b$, and the dihedral angles at edges $AC$ and $BC$ are $\alpha$ and $\beta$ respectively. Since the lateral edges of the tetrahedron are equal, its height $DO$ passes through the center $O$ of the circumcircle of the base $ABC$. The center of the circumcircle of a right triangle is the midpoint of the hypotenuse. Therefore, point $O$ is the midpoint of segment $AB$. Let $M$ and $K$ be the feet of the perpendiculars dropped from point $O$ to $BC$ and $AC$ respectively. Then $OM$ and $OK$ are the midlines of triangle $ABC$. Therefore,

$$
OM = \frac{1}{2} AC = \frac{5}{2}, \quad OK = \frac{1}{2} BC = \frac{a}{2}
$$

By the theorem of three perpendiculars, $DM \perp BC$ and $DK \perp AC$, so $OMD$ and $OKD$ are the linear angles of the dihedral angles at edges $BC$ and $AC$. According to the problem, $\angle OMD = \beta$ and $\angle OKD = \alpha$. From the right triangles $OMD$ and $OKD$, we find that

$$
OD = OM \tan \beta, \quad OD = OK \tan \alpha
$$

Multiplying these equations term by term, we get:

$$
OD^2 = OM \cdot OK \tan \alpha \tan \beta = \frac{1}{1} ab \tan \alpha \tan \beta = \frac{1}{2} S \tan \alpha \tan \beta,
$$

from which

$$
OD = \sqrt{\frac{1}{2} S \tan \alpha \tan \beta} = \frac{1}{2} \sqrt{2 S \tan \alpha \tan \beta}.
$$

Therefore,

$$
V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot OD = \frac{1}{3} \cdot \frac{1}{2} S \cdot \sqrt{2 S \tan \alpha \tan \beta} = \frac{1}{6} S \sqrt{2 S \tan \alpha \tan \beta}
$$

## Answer

$\frac{1}{6} S \sqrt{2 S \tan \alpha \tan \beta}$.