Question

How do you use the law of cosines to find AB given a triangle ABC with Angle C=41, AC=13 and CB=29?

Answer 1

`c^2 = a^2 + b^2 - 2 a b cos C`

`= 29^2+13^2- 2(29)(13) cos 41^circ` `= 1010 - 754 \ cos 41^circ`

`c approx 21.00`

Explanation:

Typically we label the opposite sides with small letters, so we have angle `C=41^circ` and `b=13` and `a=29`. The Law of Cosines is

`c^2 = a^2 + b^2 - 2 a b cos C` `= 29^2+13^2- 2(29)(13) cos 41^circ` `= 1010 - 754 cos 41^circ`

`c approx 21.00`