## Trigonometry (10th Edition)

$Adjacent$ = $2\sqrt2$ $Opposite$ = 2$\sqrt2$
Hypotenuse = 4 We must solve for the other sides. $\sin$ 45$^{\circ}$ = $\frac{Opposite}{Hypotenuse}$ $\sin$ 45$^{\circ}$ = $\frac{Opposite}{4}$ $\frac{\sqrt2}{2}$ = $\frac{Opposite}{4}$ 4($\frac{\sqrt2}{2}$) = $Opposite$ 2$\sqrt2$ = $Opposite$ $\tan$ 45$^{\circ}$ = $\frac{Opposite}{Adjacent}$ $\tan$ 45$^{\circ}$ = $\frac{2\sqrt2}{Adjacent}$ 1 = $\frac{2\sqrt2}{Adjacent}$ $Adjacent$ = $2\sqrt2$