Consider triangle ABC with sides AB, BC, and CA. The length of side AB is 10 units, and the length of side AC is 10 units as well.

Triangle ABC is thus an isosceles triangle.

If the angle opposite side BC (angle A) measures 40 degrees, what is the length of side BC?

Use the Law of Cosines, which states that for any triangle with sides $a$, $b$, and $c$, and the angle $ heta$ opposite side $c$: $$c^2 = a^2 + b^2 - 2ab \ cos( heta).$$