Answer
$\cos A=\frac{2\sqrt 5}{5}$
$\sin A=\frac{\sqrt 5}{5}$
$\tan A=\frac{1}{2}$
Work Step by Step
Using the Pythagorean theorem to find the other side of the right triangle:
$a^2+b^2=c^2$
$6^2+12^2=c^2$
$c=6\sqrt 5$
$\cos A=\frac{12}{6\sqrt 5}=\frac{2\sqrt 5}{5}$
$\sin A=\frac{6}{6\sqrt 5}=\frac{\sqrt 5}{5}$
$\tan A=\frac{6}{12}=\frac{1}{2}$
