# In triangle ABC, a=5, b=12, c=13, how do you find the cosine of each of the angles?

Jul 22, 2017

Since ${5}^{2} + {12}^{2} = {13}^{2}$, this is a right triangle with hypotenuse $c = 13$.
$\cos A = \frac{b}{c} = \frac{12}{13}$
$\cos B = \frac{a}{c} = \frac{5}{13}$
$\cos C = \cos {90}^{\circ} = 0$