## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 8 - Review Exercises - Page 410: 51

#### Answer

$(-1, \sqrt{3})$ is equivalent to $(2, 120^\circ)$ in polar coordinates.

#### Work Step by Step

For $(-1, \sqrt{3})$, $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$ $tan\theta = \frac{\sqrt{3}}{-1} = -\sqrt{3}$ $\theta = arctan (-\sqrt{3}) = 120^\circ$ (As $(-1, \sqrt{3})$ is in the second quadrant) Hence, $(-1, \sqrt{3})$ is equivalent to $(2, 120^\circ)$ in polar coordinates.

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