Answer
$-60^o$
Work Step by Step
The value of $\theta$ must be in the interval $[-90^o, 90^o]$.
Note that $\sin{60^o} = \frac{\sqrt{3}}{2}$.
Since $\sin{-\theta} = -\sin{\theta}$,
Then $\sin{(-60^o)}=-\sin{60^o} = -\frac{\sqrt{3}}{2}$
Thus, $\arcsin{(-\frac{\sqrt3}{2})}=-60^o$
