Triangle $OAB$ has $O=(0,0)$, $B=(5,0)$, and $A$ in the first quadrant. In addition, $\angle ABO=90^\circ$ and $\angle AOB=30^\circ$. Suppose that $OA$ is rotated $90^\circ$ counterclockwise about $O$. What are the coordinates of the image of $A$? 
$\mathrm{(A)}\ \left( - \frac {10}{3}\sqrt {3},5\right) \qquad \mathrm{(B)}\ \left( - \frac {5}{3}\sqrt {3},5\right) \qquad \mathrm{(C)}\ \left(\sqrt {3},5\right) \qquad \mathrm{(D)}\ \left(\frac {5}{3}\sqrt {3},5\right) \qquad \mathrm{(E)}\ \left(\frac {10}{3}\sqrt {3},5\right)$
Since $\angle ABO=90^\circ$, and $\angle AOB=30^\circ$, we know that this triangle is one of the [Special Right Triangles](https://artofproblemsolving.com/wiki/index.php/Special_Right_Triangles).
We also know that $B$ is $(5,0)$, so $B$ lies on the x-axis. Therefore, $OB = 5$.
Then, since we know that this is a Special Right Triangle ($30$-$60$-$90$ triangle), we can use the proportion \[\frac{5}{\sqrt 3}=\frac{AB}{1}\] to find $AB$.
We find that \[AB=\frac{5\sqrt 3}{3}\]
That means the coordinates of $A$ are $\left(5,\frac{5\sqrt 3}3\right)$.
Rotate this triangle $90^\circ$ counterclockwise around $O$, and you will find that $A$ will end up in the second quadrant with the coordinates
$\boxed{ \left( -\frac{5\sqrt 3}3, 5\right)}$.
Note: To better visualize this, one can sketch a diagram.