Triangle $OAB$ has $O=(0,0)$, $B=(9,0)$, and $A$ in the first quadrant, with $\angle ABO=90^\circ$ and $\angle AOB=30^\circ$. Suppose $OA$ is rotated $90^\circ$ counterclockwise about $O$. What are the coordinates of the image of $A$?

1. Since $\triangle OAB$ is $30^\circ$-$60^\circ$-$90^\circ$, $A=\left(9,\frac{9\sqrt3}3\right)=(9,3\sqrt3)$, by theorem 23.
2. Rotating $90^\circ$ CCW: $(x,y)\mapsto(-y,x)$, so $A'=(-3\sqrt3,9)$.
3. The answer is $\boxed{(-3\sqrt3,9)}$.