12.391 The sides of the parallelogram are equal to $a$ and $b(a<b)$. The smaller diagonal forms an obtuse angle with the smaller side and an angle $\alpha$ with the larger side. Find the larger diagonal of the parallelogram.
## Solution.

B

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-282.jpg?height=197&width=368&top_left_y=532&top_left_x=67)

Let in the given parallelogram $ABCD$, $a=AB, b=BC, \alpha=\angle ADB, \beta=\angle BDC$. By the Law of Sines for $\triangle ABD$,

$$
\begin{aligned}
& \frac{a}{\sin \alpha}=\frac{b}{\sin \beta} \Rightarrow \sin \beta=\frac{b}{a} \sin \alpha \\
& \cos \beta=-\frac{\sqrt{a^{2}-b^{2} \sin ^{2} \alpha}}{a}\left(\beta>90^{\circ}\right) \Rightarrow
\end{aligned}
$$

$\Rightarrow$ by the Law of Cosines

$AC^{2}=a^{2}+b^{2}-2ab \cdot \cos (\alpha+\beta)=a^{2}+b^{2}+2b\left(\cos \alpha \cdot \sqrt{a^{2}-b^{2} \sin ^{2} \alpha}+b \sin ^{2} \alpha\right)$.

Answer: $\sqrt{a^{2}+b^{2}+2b\left(\cos \alpha \cdot \sqrt{a^{2}-b^{2} \sin ^{2} \alpha}+b \cdot \sin ^{2} \alpha\right)}$.