[ Law of Cosines ]

One side of a triangle is equal to 6, the second side is equal to $2 \sqrt{7}$, and the angle opposite to it is $60^{\circ}$. Find the third side of the triangle.

#
Using the cosine theorem, write an equation for the unknown side.

## Solution

Let $x$ be the third side of the triangle. Then, by the cosine theorem,

$$
(2 \sqrt{7})^{2}=x^{2}+6^{2}-6 x, \text{ or } x^{2}-6 x+8=0 \text{. }
$$

From this equation, we find that $x=2$ or $x=4$.

Let $x$ be the third side of the triangle. Then, by the cosine theorem,

$$
(2 \sqrt{7})^{2}=x^{2}+6^{2}-6 x \text{, or } x^{2}-6 x+8=0 \text{. }
$$

From this equation, we find that $x=2$ or $x=4$.

Let $x$ be the third side of the triangle. Then, by the cosine theorem,

$$
(2 \sqrt{7})^{2}=x^{2}+6^{2}-6 x \text{, or } x^{2}-6 x+8=0 .
$$

From this equation, we find that $x=2$ or $x=4$.

## Answer

2 or 4.