Question

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 $\overline{OA}$ is rotated $90^\circ$ counterclockwise about $O.$ What are the coordinates of the image of $A?$

$\left(-\dfrac{10}{3}\sqrt{3},\,5\right)$

$\left(-\dfrac{5}{3}\sqrt{3},\,5\right)$

$\left(\sqrt{3},\,5\right)$

$\left(\dfrac{5}{3}\sqrt{3},\,5\right)$

$\left(\dfrac{10}{3}\sqrt{3},\,5\right)$

Solution:

Because $\angle ABO=90^\circ,$ segment $AB$ is vertical, so $A=\left(5,\,5\tan 30^\circ\right)=\left(5,\,\tfrac{5\sqrt3}{3}\right).$

A $90^\circ$ counterclockwise rotation about the origin sends $(x,y)$ to $(-y,x),$ so the image of $A$ is $\left(-\tfrac{5\sqrt3}{3},\,5\right).$

Thus, the correct answer is B.

2008 AMC 10B Problem 14

Solution:

Problem 14 in Other Years