Finding Side Lengths in a Triangle Using the Law of Cosines

Very Hard Geometry Triangles

In triangle ABC, the lengths of the sides are given by the following relationships:

Side AB = 4 units, Side AC = 7 units, and the angle A between sides AB and AC is measured at 120 degrees. Determine the length of side BC using the Law of Cosines, which is expressed by the formula: $$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$ where c is the side opposite angle C, and a and b are the other two sides.

$$BC = \sqrt{4^2 + 7^2 - 2 \cdot 4 \cdot 7 \cdot ( -\frac{1}{2})}$$

$$BC = \sqrt{4^2 + 7^2 - 2 \cdot 4 \cdot 7 \cdot (\frac{1}{2})}$$

$$BC = \sqrt{4^2 + 7^2 + 2 \cdot 4 \cdot 7 \cdot (\frac{1}{2})}$$

$$BC = \sqrt{4^2 + 7^2 + 2 \cdot 4 \cdot 7 \cdot ( -\frac{1}{2})}$$

Hint

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