12.059. Find the acute angle of the rhombus, knowing that the volumes of the bodies obtained by rotating the rhombus around its larger diagonal and around its side are in the ratio of $1: 2 \sqrt{5}$.
## Solution.

Let the side of the rhombus be $a$, and the acute angle be $\alpha$ (Fig. 12.61a). This means that $\angle DAB = \alpha$, and $AC$ is the larger diagonal. When the rhombus rotates around the larger diagonal, a body consisting of 2 equal cones is formed (Fig. 12.61b). The volume of each cone $V' = \frac{1}{3} S_{\text{base}}' \cdot h'$, where $h'$ is the height of the cone, equal to $AO$, and $S_{\text{base}} = \pi r^2$, where $r = BO$. From $\triangle ABO$, we have $AO = a \cos \frac{\alpha}{2}$, $BO = a \sin \frac{\alpha}{2}$. Therefore, $V_1 = \frac{2}{3} \pi \left(a^2 \sin \frac{\alpha}{2}\right) a^2 \cos \frac{\alpha}{2} = \frac{2 \pi}{3} a^3 \sin^2 \frac{\alpha}{2} \cos \frac{\alpha}{2}$. The volume of the body $V_2$, obtained by rotating the rhombus $ABCD$ around the side $DC$, can be found as the volume of a cylinder with the centers of the circles at the bases $K$ and $L$ (since the cones with vertices $C$ and $D$ are equal). $V_2 = S_{\text{base}}'' \cdot h''$, where $h'' = KL = CD = a$. The area of the base $S_{\text{base}} = \pi AL^2$; since $AC$ is the diagonal of the rhombus, $AC = 2a \cos \frac{\alpha}{2}$, and from $\triangle ALC$, $AL = AC \cdot \sin \frac{\alpha}{2} = 2a \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$. Then $S_{\text{base}} = \pi \cdot 4a^2 \sin^2 \frac{\alpha}{2} \cdot \cos^2 \frac{\alpha}{2} = \pi a^2 \sin^2 \alpha$. Therefore, the volume $V_2 = \pi a^3 \sin^2 \alpha$.

According to the condition $V_1 : V_2 = 1 : 2\sqrt{5}$, this means that

\[
\frac{2 \pi a^3 \sin^2 \frac{\alpha}{2} \cos \frac{\alpha}{2}}{3 \cdot 4 \pi a^3 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}} = \frac{1}{2 \sqrt{5}},
\]
from which $3 \cos \frac{\alpha}{2} = \sqrt{5}$. Therefore, $\cos \frac{\alpha}{2} = \frac{\sqrt{5}}{3}$, $\cos \alpha = 2 \cos^2 \frac{\alpha}{2} - 1 = 2 \cdot \frac{5}{9} - 1 = \frac{1}{9}$. Therefore, $\alpha = \arccos \frac{1}{9}$.

Answer: $\arccos \frac{1}{9}$.