Answer
$\theta = 60^{\circ}$
$\theta = 120^{\circ}$
Work Step by Step
$sin~\theta = \frac{\sqrt{3}}{2}$
$\frac{y}{r} = \frac{\sqrt{3}}{2}$
We can let $y = \sqrt{3}$ and $r = 2$.
Then: $x^2 = r^2-y^2$
$x = \pm \sqrt{r^2-y^2}$
$x = \pm \sqrt{(2)^2-(\sqrt{3})^2}$
$x = 1$ or $x = -1$
If $x = 1$ and $y = \sqrt{3}$, then $\theta$ makes an angle of $60^{\circ}$ above the positive x-axis. Then $\theta = 60^{\circ}$
If $x = -1$ and $y = \sqrt{3}$, then $\theta$ makes an angle of $60^{\circ}$ above the negative x-axis. Then $\theta = 180^{\circ}-60^{\circ} = 120^{\circ}$
