12.076. 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 a side of the base and the midpoint of the opposite lateral edge. Find the angle between this plane and the plane of the base.
Solution.

Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a right prism, $A B C D$ be a rhombus, $\angle B A D=\alpha$ ( $\alpha<90^{\circ}$ ), $A A_{1}: A B=k, C_{1} E=E C, A D E F$ be a section of the prism (Fig. 12.78); we need to find $\angle(A B C) ;(F A D)$. 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 $\angle F K B=\operatorname{arctg} \frac{k}{2 \sin \alpha}$.

Answer: $\operatorname{arctg} \frac{k}{2 \sin \alpha}$.