3.74. A plane is drawn through a side of a rhombus, forming angles $\alpha$ and $2 \alpha$ with the diagonals. Find the acute angle of the rhombus.
3.74. Given that $ABCD$ is a rhombus, $AD \in \beta$, $\angle A$ is acute; the diagonals of the rhombus form angles $\alpha$ and $2\alpha$ with $\beta$ (Fig. 3.75). Let $O$ be the point of intersection of the diagonals. Draw $OO_1 \perp \beta$; thus, $\sin \angle OAO_1 = \frac{OO_1}{OA}$ and $\sin \angle ODO_1 = \frac{OO_1}{OD}$. Since $\angle A$ is acute, $AC > BD$, $\sin \angle OAO_1 < \sin \angle ODO_1$, i.e., $\angle OAO_1 = \alpha$, and $\angle ODO_1 = 2\alpha$. Let $OO_1 = a$; then $OA = \frac{a}{\sin \alpha}$, and $OD = \frac{a}{\sin 2\alpha}$. In $\triangle AOD$, we have

$$
\operatorname{ctg} \frac{A}{2} = \frac{OA}{OD} = \frac{\sin 2\alpha}{\sin \alpha} = 2 \cos \alpha
$$

from which $\frac{A}{2} = \operatorname{arcctg}(2 \cos \alpha)$, i.e., $A = 2 \operatorname{arcctg}(2 \cos \alpha)$.