Answer
$y=-\frac{1}{2}$
Work Step by Step
We know that $x^2+y^2=1$.
Substitute $\frac{\sqrt3}{2}$ to $x$ to obtain:
$(\frac{\sqrt3}{2})^2+y^2=1\\
y^2=1-(\frac{\sqrt3}{2})^2\\
y^2=1-\frac{3}{4}\\
y^2=\frac{1}{4}\\
\sqrt{y^2}=\pm \sqrt{\frac{1}{4}}\\
y=\pm \frac{1}{2}$
Since the point is in the fourth quadrant, then $y$ must be negative.
Thus, $y=-\frac{1}{2}.$
