Problem 127

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Problem 127: (2008 AMC 10B, 14)

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)$

Solution

Exercise 127: Solution

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 $A$?

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