3.32. At the base of a right prism lies a rhombus with an acute angle $\alpha$. The ratio of the height of the prism to the side of the base is $k$. A plane is drawn through the side of the base and the midpoint of the opposite lateral edge. Find the angle between this plane and the plane of the base.
3.32. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a right prism, $A B C D$ is a rhombus, $\angle B A D=\alpha\left(\alpha<90^{\circ}\right), A A_{1}: A B=k, C_{1} E=E C, A D E F$ is a section of the prism (Fig. 3.34); it is required to find the angle between the section and the base $(A B C)$; (FAD). Let $A B=a$; then $A A_{1}=k a$. Draw $B K \perp A D$; then $F K \perp A D$ (by the theorem of three perpendiculars) and $\angle F K B$ is the linear angle between the section and the base $A B C D$. From $\triangle A K B$ we find $B K=a \sin \alpha ;$ since $E F \| A D$, then $F B=\frac{1}{2} B B_{1}=\frac{1}{2} k a$. Finally, from $\triangle F B K$ we get $\operatorname{tg} \angle F K B=\frac{F B}{B K}=$ $=\frac{k a}{2 a \sin \alpha}=\frac{k}{2 \sin \alpha}$, hence $\angle F K B=\operatorname{arctg} \frac{k}{2 \sin \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-519.jpg?height=529&width=654&top_left_y=107&top_left_x=256)

Fig. 3.34

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-519.jpg?height=506&width=515&top_left_y=1135&top_left_x=58)

Fig. 3.35