If the angles A, B of triangle ABC are such that $\displaystyle cos A = {-4/5}$ and $\displaystyle cos B = {12/13}$, find without using tables the value of Cos C'
Thanks.
$\displaystyle C = \pi - A - B$
$\displaystyle \cos C = \cos(\pi - A - B)$
Using identities:
$\displaystyle \cos C = \cos(\pi - A)\cos B + \sin(\pi - A)\sin B$
$\displaystyle \cos C = -\cos A \cos B + \sin A \sin B$
You should be able to calculate $\displaystyle \sin A$ and $\displaystyle \sin B$.