[ Law of Sines ] [ Area of a Triangle (using two sides and the included angle).]

In parallelogram $A B C D(A B \| C D)$, the diagonals $A C=c, B D=\frac{\sqrt{3}}{\sqrt{2}} c$. Find the area of the parallelogram if $\angle C A B=60^{\circ}$.
Apply the Law of Sines to triangle $C O D (O$ - the intersection point of the diagonals of the parallelogram) and use the formula

$$
S_{\mathrm{ABCD}}=\frac{1}{2} \cdot A C \cdot B D \cdot \sin \angle C O D
$$

## Solution

Let the diagonals of the parallelogram intersect at point $O$. Denote $\angle C O D=\alpha$. Consider triangle $C O D$. In it,

$$
\begin{gathered}
\angle O C D=\angle C A B=60^{\circ}, \angle C O D=\alpha, \angle O D C=120^{\circ}-\alpha \\
O C=\frac{1}{2} A C=\frac{c}{2}, O D=\frac{1}{2} B D=\frac{\sqrt{3}}{2 \sqrt{2}} c .
\end{gathered}
$$

By the Law of Sines,

$$
\frac{O D}{\sin 60^{\circ}}=\frac{O C}{\sin \left(120^{\circ}-\alpha\right)} \text {, or } \frac{\frac{\sqrt{3}}{2 \sqrt{2}} c}{\frac{\sqrt{3}}{2}}=\frac{\frac{c}{2}}{\sin \left(120^{\circ}-\alpha\right)}
$$

from which we find that $\sin \left(120^{\circ}-\alpha\right)=\frac{\sqrt{2}}{2}$. Clearly, $120^{\circ}-\alpha \neq 135^{\circ}$, so $120^{\circ}-\alpha=45^{\circ}$. Then $\alpha=120^{\circ}-45^{\circ}$

$=75^{\circ}$. Therefore,

$$
S_{\mathrm{ABCD}}=\frac{1}{2} \cdot A C \cdot B D \cdot \sin \alpha=\frac{1}{2} \cdot c \cdot \frac{\sqrt{3}}{\sqrt{2}} c \cdot \frac{\sqrt{6}+\sqrt{2}}{4}=\frac{c^{2}}{8}(3+\sqrt{3}) .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-30.jpg?height=517&width=504&top_left_y=1365&top_left_x=777)

## Answer

$S=\frac{c^{2}}{8}(3+\sqrt{3})$