## Algebra 1: Common Core (15th Edition)

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