Answer
$\tan\theta=-\frac{\sqrt 3}{3}$
Work Step by Step
1.We get this information from the sine value
$y=2$
$r=2$
2. We can use y and r to find x (we need to remember that sign of x depends on quadrant, since it ends in quadrant II, x must be negative)
$r=\sqrt {y^{2}+x^{2}}$
$y=-\sqrt {r^{2}-x^{2}}=-\sqrt {2^{2}-1^{2}}=-\sqrt {3}$
4. Then insert values,
$\tan\theta=\frac{-\sqrt 3}{3}$
