12.160. The ratio of the perimeter of a rhombus to the sum of its diagonals is $k$. Find the angles of the rhombus and the permissible values of $k$.
## Solution.

Let the side of the rhombus $ABCD$ be $a$, and the acute angle $\alpha$ (Fig. 12.32). Then the diagonals of the rhombus are $2 a \sin \frac{\alpha}{2}$ and $2 a \cos \frac{\alpha}{2}$.

## According to the condition

$$
\begin{aligned}
& k=\frac{4 a}{2 a \sin \frac{\alpha}{2}+2 a \cos \frac{\alpha}{2}}=\frac{2}{\sin \frac{\alpha}{2}+\cos \frac{\alpha}{2}}= \\
& =\frac{2}{\sqrt{\left(\sin \frac{\alpha}{2}+\cos \frac{\alpha}{2}\right)^{2}}}=\frac{2}{\sqrt{1+\sin \alpha}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0916.jpg?height=672&width=455&top_left_y=556&top_left_x=806)

Fig. 12.32

Since $0<\alpha<\frac{\pi}{2}$, then $1<1+\sin \alpha \leq 2, \sqrt{2} \leq \frac{2}{\sqrt{1+\sin \alpha}}<2, \sqrt{2} \leq k<2$.

Thus, $k^{2}=\frac{4}{1+\sin \alpha} ; \sin \alpha=\frac{4-k^{2}}{k^{2}} ; \alpha=\arcsin \frac{4-k^{2}}{k^{2}}$.

Answer: $\arcsin \frac{4-k^{2}}{k^{2}}, \pi-\arcsin \frac{4-k^{2}}{k^{2}} ; \sqrt{2} \leq k<2$.