2008 AMC 10B Problems/Problem 14

Problem

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 $A$ ?

As $\angle ABO=90^\circ$ and $A$ in the first quadrant, we know that the $x$ coordinate of $A$ is $5$ . We now need to pick a positive $y$ coordinate for $A$ so that we'll have $\angle AOB=30^\circ$ .

Substituting into the previous equation, we get $AB^2 = \frac{25}3$ , hence $AB=\frac{5\sqrt 3}3$ .

This means that the coordinates of $A$ are $\left(5,\frac{5\sqrt 3}3\right)$ .

After we rotate $A$ $90^\circ$ counterclockwise about $O$ , it will get into the second quadrant and have the coordinates $\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }$ .