12.304. The diagonal of a rectangular parallelepiped is equal to $l$ and forms angles $\alpha$ and $\beta$ with two adjacent faces. Find the volume of the parallelepiped.
## Solution.

Let in a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1} B_{1} D=l$ (Fig. 12.166).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1059.jpg?height=726&width=770&top_left_y=65&top_left_x=269)

Fig. 12.167

Since the edges $D A$ and $D C$ are perpendicular to the faces $A A_{1} B_{1} B$ and $C C_{1} B_{1} B$, then $A B_{1}$ and $B_{1} C$ are the projections of $B_{1} D$ onto these faces, $\angle A B_{1} D=\alpha, \angle C B, D=\beta$.

In $\triangle B_{1} A D\left(\angle B_{1} A D=90^{\circ}\right): A D=B_{1} D \sin \angle A B_{1} D=l \sin \alpha$.

In $\triangle B_{1} C D\left(\angle B_{1} C D=90^{\circ}\right): C D=B_{1} D \sin \angle C B_{1} D=l \sin \beta$.

In the rectangle $A B C D: B D^{2}=A D^{2}+C D^{2}=l^{2}\left(\sin ^{2} \alpha+\sin ^{2} \beta\right)$.

In $\triangle B_{1} B D\left(\angle B_{1} B D=90^{\circ}\right)$ :

$$
\begin{aligned}
& B B_{1}=\sqrt{B_{1} D^{2}-B D^{2}}=l \sqrt{1-\left(\sin ^{2} \alpha+\sin ^{2} \beta\right)}=l \sqrt{\cos ^{2} \alpha-\sin ^{2} \beta}= \\
& =l \sqrt{\frac{1+\cos 2 \alpha}{2}-\frac{1-\cos 2 \beta}{2}}=l \sqrt{\frac{\cos 2 \alpha+\cos 2 \beta}{2}}=l \sqrt{\cos (\alpha+\beta) \cos (\alpha-\beta)}
\end{aligned}
$$

The volume of the rectangular parallelepiped $V=A D \cdot C D \cdot B B_{1}=$ $=l^{3} \sin \alpha \sin \beta \sqrt{\cos (\alpha+\beta) \cos (\alpha-\beta)}$.

Answer: $l^{3} \sin \alpha \sin \beta \sqrt{\cos (\alpha+\beta) \cos (\alpha-\beta)}$.