12.292. 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.
Solution.

Let $ABCD$ be a rhombus, through the side of which a plane $\pi$ is drawn (Fig. 12.154), $O$ - the point of intersection of the diagonals of the rhombus.

Draw perpendiculars $BB_1$ and $CC_1$ to the plane $\pi$, $\angle CAC_1 = \alpha$, $\angle BDB = 2\alpha$.

$BC \| AD$, and $AD$ belongs to the plane $\pi$. Thus, the line $BC$ is parallel to the plane $\pi$ and $BB_1 = CC_1$.

In the right triangles $AC_1C$ and $DB_1B$ with equal legs $CC_1$ and $BB_1$, $\angle CAC_1 = \alpha$ and $\angle BDB = 2\alpha \Rightarrow \angle BAD$ is the acute angle of the rhombus.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1046.jpg?height=753&width=846&top_left_y=68&top_left_x=271)

Fig. 12.155

In $\triangle AC_1C (\angle AC_1C = 90^\circ): AC = \frac{CC_1}{\sin \alpha}$.

In $\triangle DB_1B (\angle DB_1B = 90^\circ): BD = \frac{BB_1}{\sin 2\alpha}$.

In $\triangle AOD (\angle AOD = 90^\circ): \operatorname{ctg} \angle OAD = \frac{AO}{OD} = \frac{\frac{1}{2} AC}{\frac{1}{2} BD} = \frac{CC_1}{\sin \alpha} : \frac{BB_1}{\sin 2\alpha} = 2 \cos \alpha \Rightarrow \angle OAD = \operatorname{arctg}(2 \cos \alpha)$

$\angle BAD = 2 \angle OAD = 2 \operatorname{arcctg}(2 \cos \alpha)$.

Answer: $2 \operatorname{arcctg}(2 \cos \alpha)$.