3.100. 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.
3.100. By the condition, $S A B C D-$ is a pyramid, $A B C D$ is a rectangle, $S E \perp(A B C), \angle S A E=\angle S B E=$ $=\angle S C E=\angle S D E=\beta$, from which it follows that $E$ is the point of intersection of the diagonals $A C$ and $B D$ (Fig. 3.100), $\angle A E B=\alpha$ and $O$ is the center of the sphere circumscribed around the pyramid, which lies on $S E, R$ is the radius of this sphere. Since $\angle E S D=90^{\circ}-\beta$ and $O S=O D=R$, then $\angle O D E=\angle E S D=$ $=90^{\circ}-\beta$. Therefore, $\angle E O D=2 \angle E S D=$ $=180^{\circ}-2 \beta$ as the external angle of $\triangle S O D$. From $\triangle O E D$ we find $E D=R \sin \left(180^{\circ}-2 \beta\right)=R \sin 2 \beta$, hence $B D=2 R \sin 2 \beta$. From $\triangle S E D$ we obtain

$$
S E=E D \operatorname{tg} \beta=R \sin 2 \beta \operatorname{tg} \beta=2 R \sin ^{2} \beta
$$

Finally, we have

$$
\begin{gathered}
V=\frac{1}{3} S_{A B C D} \cdot S F=\frac{1}{3} \cdot \frac{1}{2} B D^{2} \sin \alpha \cdot 2 R \sin ^{2} \beta= \\
=\frac{i}{6} \cdot 8 R^{3} \sin ^{2} 2 \beta \sin ^{2} \beta \sin \alpha=\frac{4}{3} R^{3} \sin ^{2} 2 \beta \sin ^{2} \beta \sin \alpha
\end{gathered}
$$