## Precalculus (6th Edition) Blitzer

$13^o$
RECALL: An angle $\theta$, where $360^o \lt \theta \lt 720^o$, is coterminal with: $\theta - 360^o$ Thus, the given angle is coterminal with: $=553^o - 360^o \\=193^o$ $553^o$ is coterminal with $193^o$. $193^o$ is in Quadrant III so $553^o$ is also in Quadrant III. RECALL: The following are the means on how to find the reference angle of an angle $0\leq \theta \lt 360^\circ$ based on its position: (1) Quadrant I: $\theta$ (itself) (2) Quadrant II: $180^o-\theta$ (3) Quadrant III: $\theta - 180^o$ (4) Quadrant IV: $360^o-\theta$ Use formula (3) above to obtain: reference angle of $553^o$ = reference angle of $193^o$, which is $193^o - 180^o = 13^o$