12.335. The base of the pyramid is a rectangle, where the angle between the diagonals is $\alpha$. A sphere of radius $R$ is circumscribed around this pyramid. Find the volume of the pyramid if all its lateral edges form an angle $\beta$ with the base.
Solution.

Let $EO$ be the height of the pyramid $EABCD$, and the rectangle $ABCD$ be its base (Fig. 12.196). Since all lateral edges form the same angle with the plane of the base, point $O$ is the center of the circle circumscribed around the rectangle $ABCD$, the point of intersection of its diagonals, and the center of the sphere circumscribed around the pyramid lies on the ray $EO$. The triangle $\triangle AEC$ is inscribed in the great circle of the sphere. Then $AC = 2R \sin \angle AEC = 2R \sin (180^\circ - \beta) = 2R \sin 2\beta, OA = R \sin 2\beta$.

The area of the base of the pyramid $S = \frac{1}{2} AC^2 \sin \alpha = 2R^2 \sin^2 2\beta \sin \alpha$.

From $\triangle AOE (\angle AOE = 90^\circ): EO = AO \operatorname{tg} \angle EAO = R \sin 2\beta \operatorname{tg} \beta$.

The volume of the pyramid:

$$
V = \frac{1}{3} S \cdot EO = \frac{2}{3} R^3 \sin^3 2\beta \sin \alpha \operatorname{tg} \beta = \frac{4}{3} R^3 \sin^2 2\beta \sin^2 \beta \sin \alpha
$$

Answer: $\frac{4}{3} R^3 \sin^2 2\beta \sin^2 \beta \sin \alpha$.