[ Law of Cosines]

In a triangle, the base is 12; one of the angles at the base is $120^{\circ}$; the side opposite this angle is 28.

Find the third side.
Apply the Law of Cosines.

## Solution

Let in triangle $ABC$ it is known that $\angle C=120^{\circ}, AC=12, AB=28$. Denote $BC=x$. By the Law of Cosines,

$$
AB^{2}=AC^{2}+BC^{2}-2 AC \cdot BC \cos \angle C
$$

or

$$
784=x^{2}+144+12 x, x^{2}+12 x-640=0
$$

The positive root of this equation is $x=20$

![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-12.jpg?height=263&width=730&top_left_y=1809&top_left_x=664)

## Answer

20. 

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