$\frac{\text { [ Angles between lines and planes }]}{[\text { Rectangular parallelepipeds }]}$

Given a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$. A plane is drawn through the line $B D 1$, parallel to the line $A C$. Find the angle between this plane and the base plane of the parallelepiped, if $A B=a, B C$ $=b, C C 1=c$.
The line $A 1 C 1$ is parallel to the line $A C$. The plane $A 1 B 1 C 1 D 1$ passes through the line $A 1 C 1$, which is parallel to the intersecting plane, and has a common point $D 1$ with the intersecting plane, so the line $a$ of intersection of the intersecting plane with the plane $A 1 B 1 C 1 D 1$ is parallel to the line $A 1 C 1$. Let $B 1 M$ be the perpendicular dropped from the vertex $B 1$ to the line $a$. Then $B 1 M$ is the orthogonal projection of the oblique $B M$ onto the plane $A 1 B 1 C 1 D 1$. By the theorem of three perpendiculars, $B M \perp a$, so $B M B 1$ is the linear angle of the dihedral angle formed by the intersecting plane and the plane $A 1 B 1 C 1 D 1$. Let $\angle B M B 1=\alpha$. The segment $B 1 M$ is twice the height $B 1 H$ of the right triangle $A 1 B 1 C 1$, drawn from the vertex of the right angle, so

![](https://cdn.mathpix.com/cropped/2024_05_06_2c7701213788eaf04894g-05.jpg?height=87&width=662&top_left_y=0&top_left_x=698)

From the right triangle $B M B 1$ we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_2c7701213788eaf04894g-05.jpg?height=119&width=618&top_left_y=146&top_left_x=717)

## Answer

$\operatorname{arctg} \frac{\sqrt{\alpha^{2}+b^{2}}}{2 \omega b}$.