12.437 The base of a right parallelepiped is a parallelogram with diagonals $a$ and $b(a>b)$ and an acute angle $\alpha$ between them. The smallest diagonal of the parallelepiped forms an acute angle $\beta$ with the larger diagonal of the base. Find the volume of the parallelepiped.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-310.jpg?height=300&width=410&top_left_y=703&top_left_x=101)

Let \( A B C D A_{1} B_{1} C_{1} D_{1} \) be the given right parallelepiped and \( A C_{1} \) be its smaller diagonal, \( A C = b, \quad B D = a \). In the plane of the base \( A B C D \), draw a line \( l \parallel B D \) through \( A \) and drop a perpendicular \( C_{1} K \) from vertex \( C_{1} \) to line \( l \). Then \( l \perp C K \) (Theorem 12), \( \angle C A K = \alpha \), \( \angle C_{1} A K = \beta \) (by condition) \( \Rightarrow \). \( \Rightarrow C_{1} K = A C_{1} \cdot \sin \beta, C K = b \cdot \sin \alpha \).

But \( C_{1} C^{2} = C_{1} K^{2} - C K^{2}, A C_{1}^{2} = b^{2} + C_{1} C^{2} \Rightarrow C_{1} K^{2} = \left(b^{2} + C_{1} C^{2}\right) \cdot \sin^{2} \beta \Rightarrow \) \( \Rightarrow C_{1} C^{2} = \left(b^{2} + C_{1} C^{2}\right) \cdot \sin^{2} \beta - b^{2} \sin^{2} \alpha \Rightarrow C_{1} C = \frac{b}{\cos \beta} \sqrt{\sin^{2} \beta - \sin^{2} \alpha} = \) \( = \frac{b}{\cos \beta} \sqrt{\sin (\beta - \alpha) \cdot \sin (\beta + \alpha)} \). Since \( S_{A B C D} = \frac{1}{2} a b \cdot \sin \alpha \), we get the answer.

Answer: \( V_{T} = \frac{a b^{2} \sin \alpha}{2 \cos \beta} \cdot \sqrt{\sin (\beta + \alpha) \cdot \sin (\beta - \alpha)} \).