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.

Printable View

- Feb 20th 2010, 03:47 PMdeltaxraysums and differences of angles
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. - Feb 20th 2010, 03:53 PMicemanfan
$\displaystyle C = \pi - A - B$

$\displaystyle \cos C = \cos(\pi - A - B)$

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 $\displaystyle \sin B$.