[ Rectangular parallelepipeds ] [ Angles between lines and planes $]$

The sides of the base of a rectangular parallelepiped are equal to $a$ and $b$. The diagonal of the parallelepiped is inclined to the plane of the lateral face containing the side of the base equal to $b$, at an angle of $30^{\circ}$. Find the volume of the parallelepiped.
Let $A B C D A 1 B 1 C 1 D 1$ be the given rectangular parallelepiped with bases $A B C D, A 1 B 1 C 1 D 1$ and lateral edges $A A 1, B B 1, C C 1$ and $D D 1$, where $B C=a, C D=b$. Since $B C$ is perpendicular to the plane of the face $C C 1 D 1 D$ containing the edge $C D, C D 1$ is the orthogonal projection of the diagonal $B D 1$ onto this plane. Therefore, $\angle B D 1 C$ is the angle between the line $B D 1$ and the plane of the face $C C 1 D 1 D$. According to the problem, $\angle B D 1 C=30^{\circ}$. From the right triangles $B D 1 C, B D C$ and $B D D 1$ we find that

$$
\begin{gathered}
B D 1=2 B C=2 a, B D 2=B C 2+C D 2=a 2+b 2, \\
D D 1=\sqrt{B D^{2}-B D^{2}}=\sqrt{4 a^{2}-a^{2}-b^{2}}=\sqrt{3 a^{2}-b^{2}} .
\end{gathered}
$$

Therefore,

$$
V_{A B C D A} 1 B 1 C 1 D 1=S_{A B C D} \cdot D D 1=a b \sqrt{3 a^{2}-b^{2}}
$$

## Answer

$a b \sqrt{3 a^{2}-b^{2}}$.