## Precalculus (6th Edition) Blitzer

$N 15^{\circ} E$
The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ Our equation becomes: $90^{\circ}-75^{\circ}=15^{\circ}$ Our answer is: $N 15^{\circ} E$