## Elementary Geometry for College Students (5th Edition)

$\tan\alpha=\frac{\sqrt5}{2}$ $\tan\beta=\frac{2}{\sqrt5}$
Let $x$ be the side of the triangle with an unknown length. Using Pythagorean Theorem to find $x$. $x=\sqrt{6^{2}-4^{2}}$ $x=\sqrt{20}$ $x=2\sqrt5$ $\tan\alpha=\frac{opposite}{hypotenuse}=\frac{x}{4}=\frac{2\sqrt5}{4}=\frac{\sqrt5}{2}$ $\tan\beta=\frac{opposite}{hypotenuse}=\frac{4}{x}=\frac{4}{2\sqrt5}=\frac{2}{\sqrt5}$