# Chapter 3 - Review Exercises - Page 136: 58

$\color{blue}{s=\dfrac{2\pi}{3}}$

#### Work Step by Step

RECALL: (1) $\frac{\pi}{3}$ is a special angle and $\tan{(\frac{\pi}{3})}=\sqrt3$ (2) An angle and its reference angle either have the same trigonometric function values or they differ only in sign. The required value of $s$ is in the interval $\left[\dfrac{\pi}{2}, \pi\right]$, which is in Quadrant II. Note that the reference angle of $\dfrac{2\pi}{3}$ is $\dfrac{\pi}{3}$, and that since $\dfrac{2\pi}{3}$ is in Quadrant II, its tangent value is negative, Thus, if $\tan{s} = -\sqrt{3}$ and $s$ must be in $\left[\dfrac{\pi}{2}, \pi\right]$, then $\color{blue}{s=\dfrac{2\pi}{3}}$.

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