# In which quadrant is the terminal of an angle in standard position whose measure is (-55pi)/3?

Feb 3, 2015

The answer is: $\frac{5}{3} \pi$.

Every $2 \pi$ the angles repeat. So it is possible to add, or subtract, at one angle $2 \pi$ how many times you want.

Let's sum it until we "reach" an angle of "the first round", i.e until the angle is in $\left[0 , 2 \pi\right]$.

$- \frac{55}{3} \pi + 2 \pi = - \frac{55}{3} \pi + \frac{6}{3} \pi r = - \frac{49}{3} \pi$

$- \frac{49}{3} \pi + \frac{6}{3} \pi = - \frac{43}{3} \pi$

$- \frac{43}{3} \pi + \frac{6}{3} \pi = - \frac{37}{3} \pi$

$- \frac{37}{3} \pi + \frac{6}{3} \pi = - \frac{31}{3} \pi$

$- \frac{31}{3} \pi - \frac{6}{3} \pi = - \frac{25}{3} \pi$

$- \frac{25}{3} \pi + \frac{6}{3} \pi = - \frac{19}{3} \pi$

$- \frac{19}{3} \pi + \frac{6}{3} \pi = - \frac{13}{3} \pi$

$- \frac{13}{3} \pi + \frac{6}{3} \pi = - \frac{7}{3} \pi$

$- \frac{7}{3} \pi + \frac{6}{3} \pi = - \frac{\pi}{3}$

$- \frac{\pi}{3} + \frac{6}{3} \pi = \frac{5}{3} \pi$.