12.231. A rectangular parallelepiped is inscribed in a cylinder, with its diagonal forming angles $\alpha$ and $\beta$ with the adjacent sides of the base. Find the ratio of the volume of the parallelepiped to the volume of the cylinder.
Solution.

Let in the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.98), inscribed in a cylinder, $\angle B_{1} D A=\alpha, \angle B_{1} D C=\beta, A D=a, C D=b$, the radius of the cylinder's base $R$, the common height of the cylinder and the parallelepiped $H$, $B_{1} D=d, A B$ - the projection of $A B_{1}$ on the base plane of the parallelepiped, $A B \perp A D$. Then $A B_{1} \perp A D$.

In $\triangle B_{1} A D\left(\angle B_{1} A D=90^{\circ}\right): a=d \cos \alpha$, and since $B_{1} C \perp C D$, then from $\triangle B_{1} C D$ : $b=d \cos \beta$.

The volume of the parallelepiped $V_{1}=a b H=d^{2} \cos \alpha \cos \beta H$.

$$
B D^{2}=(2 R)^{2}=4 R^{2}=a^{2}+b^{2}=d^{2}\left(\cos ^{2} \alpha+\cos ^{2} \beta\right)
$$

The volume of the cylinder

$$
\begin{aligned}
& V_{2}=\pi R^{2} H=\frac{1}{4} \pi d^{2}\left(\cos ^{2} \alpha+\cos ^{2} \beta\right) H \Rightarrow \\
& \Rightarrow \frac{V_{1}}{V_{2}}=\frac{d^{2} \cos \alpha \cos \beta H}{\frac{1}{4} \pi d^{2}\left(\cos ^{2} \alpha+\cos ^{2} \beta\right) H}=\frac{4 \cos \alpha \cos \beta}{\pi\left(\cos ^{2} \alpha+\cos ^{2} \beta\right)} \\
& \text { Answer: } \frac{4 \cos \alpha \cos \beta}{\pi\left(\cos ^{2} \alpha+\cos ^{2} \beta\right)}
\end{aligned}
$$