Task B-3.7. In a parallelogram, the obtuse angle measures $120^{\circ}$, and the lengths of the diagonals are in the ratio $\sqrt{109}: \sqrt{39}$. In what ratio are the lengths of the sides?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-18.jpg?height=705&width=1325&top_left_y=507&top_left_x=365)

From the given data and the cosine rule, we have

$$
\begin{aligned}
\frac{e^{2}}{f^{2}} & =\frac{a^{2}+b^{2}-2 a b \cos 120^{\circ}}{a^{2}+b^{2}-2 a b \cos 60^{\circ}} \\
\frac{109}{39} & =\frac{a^{2}+b^{2}+a b}{a^{2}+b^{2}-a b}
\end{aligned}
$$

From this, after simplification, we get

$$
35 a^{2}-74 a b+35 b^{2}=0
$$

(2 points)

Then, by factoring, we obtain

$$
\begin{aligned}
& 35 a^{2}-35 a b-49 a b+35 b^{2}=0 \\
& 5 a(7 a-5 b)-7 b(7 a-5 b)=0 \\
& (7 a-5 b)(5 a-7 b)=0
\end{aligned}
$$

It follows that \(7 a-5 b=0\) or \(5 a-7 b=0\), that is,

$$
\frac{a}{b}=\frac{5}{7} \quad \text { or } \quad \frac{a}{b}=\frac{7}{5}
$$

Note: Equation (*) can be solved as a quadratic in \(a\) or \(b\), or it can be divided by \(b^{2}\) to obtain a quadratic equation in \(\frac{a}{b}\):

$$
35\left(\frac{a}{b}\right)^{2}-74 \cdot \frac{a}{b}+35=0
$$