Answer
$-\frac{4}{5}$, or (A)
Work Step by Step
In Quadrant 2, x is negative and y is positive. In the context of the unit circle, that means that $\sin(\theta)>0,
\cos(\theta) < 0$
The trigonometric identity $\sin^2(\theta) + \cos^2(\theta) = 1$ applies here. We need to solve for $\cos(\theta)$
$\sin^2(\theta) + \cos^2(\theta) = 1 = (\frac{3}{5})^2 +\cos^2(\theta)$
$\cos^2(\theta) = \frac{16}{25}, \cos(\theta) = \frac{4}{5} \ or \ -\frac{4}{5}$
As stipulated earlier, in quadrant II, cosine is negative, so $\cos(\theta) = -\frac{4}{5}$
