## Precalculus (6th Edition) Blitzer

$25^o$
RECALL: The following are the means on how to find the reference angle of an angle $\theta$ based on its position: (1) Quadrant I: $\theta$ (itself) (2) Quadrant II: $180^o-\theta$ (3) Quadrant III: $\theta - 180^o$ (4) Quadrant IV: $360^o-\theta$ $205^o$ is in Quadrant III. Thus, using forrmula (3) above gives: reference angle of $205^o$ = $205^o-180^o=25^o$