#### Answer

$\color{blue}{\dfrac{4\pi}{3}}$

#### Work Step by Step

Use the inverse tangent function of a calculator in radian mode to obtain:
$s=\tan^{-1}{(\sqrt3)} = \frac{\pi}{3}$
The angle must be in the given interval, which is actually in Quadrant III.
Note that for an angle $\theta$ in Quadrant I, then
$\tan{\theta} = \tan{(\pi+\theta)}$
Thus,
$\tan{\frac{\pi}{3}} = \tan{(\pi+\frac{\pi}{3})}=\tan{\frac{4\pi}{3}}$
Therefore, then angle must be $\color{blue}{\dfrac{4\pi}{3}}$.