Answer
$y=-\frac{1}{2}$
Work Step by Step
The equation of the unit circle is whose center is at the origin is $x^{2}+y^{2}=1$.
As the point is in quadrant $IV$, $x$ is positive and $y$ is negative.
Substitute $x=\frac{\sqrt3}{2}$ into the equation above to obtain:
\begin{align*}\left(\frac{\sqrt3}{2}\right)^2 + y^{2}&=1\\
\frac{3}{4}+y^2&=1\\
y^2&=1-\frac{3}{4}\\
y^2&=\frac{1}{4}\\
y&=\pm\sqrt{\frac{1}{4}}\\
y&=\pm\frac{1}{2}
\end{align*}
Since $y$ is negative in Quadrant $IV$, then $y=-\frac{1}{2}$.
