[ Area of a triangle (using two sides and the angle between them).] Law of Cosines

Two sides of a triangle are equal to $2 \sqrt{2}$ and 3, and the area of the triangle is 3. Find the third side.
Find the sine of the angle between the given sides.

## Solution

The area of the triangle is equal to half the product of its sides and the sine of the angle between them, i.e., $3=\frac{1}{2} \cdot 3 \cdot 2 \sqrt{2} \cdot \sin \alpha$. From this, we find that $\sin \alpha=\frac{1}{\sqrt{2}}$. Then $|\cos \alpha|=\frac{1}{\sqrt{2}}$.

There are two possible cases: $\alpha=45^{\circ}$ or $\alpha=135^{\circ}$. In each case, we will find the third side using the cosine theorem:

$$
a^{2}=(2 \sqrt{2})^{2}+3^{2} \pm 2 \cdot 2 \sqrt{2} \cdot 3 \cdot \frac{1}{\sqrt{2}}=8+9 \pm 12=17 \pm 12
$$

Therefore, $a=\sqrt{29}$ or $a=\sqrt{5}$.

## Answer

$\sqrt{29}, \sqrt{5}$.