3.51. Find the acute angle of a 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 $1: 2 \sqrt{5}$.

 untranslated part:
- Найти острый угол ромба, зная, что объемы тел, получен-
- ых от вращения ромба вокруг его большей диагонали и вокруг его стороны, относятся как $1: 2 \sqrt{5}$. 

translated part:
- Find the acute angle of a 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 $1: 2 \sqrt{5}$.
### 3.51. When the rhombus $ABCD$ is rotated around the diagonal $BD$, two cones with a common base $AC$ are formed (Fig. 3.53). The volume of this figure $V_{1}=\frac{1}{3} \pi O A^{2}(O B+O D)=$ $=\frac{1}{3} \pi O A^{2} \cdot B D$. When the same rhombus is rotated around the side $DC$, a cylinder $A B B_{1} A_{1}$ is formed, from which the cone $B B_{1} C$ is removed, but the cone $A A_{1} D$ is added. Since both specified cones have equal volumes (the radii of their bases $B M$ and $A N$ are equal, their heights $C M$ and $C N$ are also equal, which follows from the equality of triangles $B C M$ and $A D N$), the volume $V_{2}$ of the second figure of rotation is equal to the volume of the cylinder $A B B_{1} A_{1}$, i.e., $V_{2}=$ $=\pi A N^{2} \cdot A B$. By the condition,

$$
\frac{V_{1}}{V_{2}}=\frac{1}{2 \sqrt{5}} ; \frac{\frac{1}{3} \pi O A^{2} \cdot B D}{\pi A N^{2} \cdot A B}=\frac{1}{2 \sqrt{5}} ; \frac{O A^{2} \cdot B D}{3 A N^{2} \cdot A B}=\frac{1}{2 \sqrt{5}}
$$

Let the side of the rhombus be $a$, and the desired angle $A D C$ be $\alpha$. Then $\angle A D O=\frac{\alpha}{2}, O A=a \sin \frac{\alpha}{2}$ (from $\triangle A O D$), $B D=2 D O=2 a \cos \frac{\alpha}{2}$, $A N=a \sin \alpha$. Therefore,

$$
\begin{gathered}
\frac{a^{2} \sin ^{2} \frac{\alpha}{2} \cdot 2 a \cos \frac{\alpha}{2}}{3 a^{2} \sin ^{2} \alpha \cdot a}=\frac{1}{2 \sqrt{5}} ; \frac{2 \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}}{12 \sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}}=\frac{1}{2 \sqrt{5}} \\
3 \cos \frac{\alpha}{2}=\sqrt{5} ; \cos \frac{\alpha}{2}=\frac{\sqrt{5}}{3}
\end{gathered}
$$

Finally, considering that $\cos \alpha=2 \cos ^{2} \frac{\alpha}{2}-1$, we get $\cos \alpha=\frac{10}{9}-1=\frac{1}{9}$, from which $\alpha=\arccos \frac{1}{9}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-533.jpg?height=512&width=404&top_left_y=367&top_left_x=44)

Fig. 3.55