Question

The law of cosines is a2 +b2-2abcosC=c2 find the value of 2abcosC

Answer 1

Since you have not provided the diagram, I cannot give a precise answer. However, I'll tell you how to solve this question.

We are given that:
a² + b² - 2abcosC = c²

We want to solve for 2abcosC. This means that we want to isolate the 2abcosC on one side of the equation.
We can do this as follows:
a² + b² - 2abcosC + 2abcosC = c² + 2abcosC
a² + b² - c² = c² + 2abcosC - c²
2abcosC = a² + b² - c²

Now in the diagram you have:
c is the side opposite to angle C
a is the side opposite to angle A
b is the side opposite to angle B

Plug in the values in the above equation, and you will simply have your answer.

Hope this helps :)

Answer 2

$\bf a^2+b^2-(2ab)cos(C)=c^2\implies a^2+b^2-c^2=(2ab)cos(C)$