#### Answer

$\dfrac{5\pi}{6}$

#### Work Step by Step

Use the inverse sine function of a calculator in radian mode to obtain:
$s=\sin^{-1}{(\frac{1}{2})} = \frac{\pi}{6}$
The angle must be in the given interval, which is actually in Quadrant II.
Note that for an angle $\theta$ in Quadrant I, then
$\sin{\theta} = \sin{(\pi-\theta)}$
Thus,
$\sin{\frac{\pi}{6}} = \sin{(\pi-\frac{\pi}{6})}=\sin{\frac{5\pi}{6}}$
Therefore, then angle must be $\color{blue}{\dfrac{5\pi}{6}}$.