Precalculus (6th Edition)

$\dfrac{5\pi}{6}$
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 $\color{blue}{\dfrac{5\pi}{6}}$.