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Revision as of 11:27, 4 July 2013

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 $A$?

$\mathrm{(A)}\ \left( - \frac {10}{3}\sqrt {3},5\right) \qquad \mathrm{(B)}\ \left( - \frac {5}{3}\sqrt {3},5\right) \qquad \mathrm{(C)}\ \left(\sqrt {3},5\right) \qquad \mathrm{(D)}\ \left(\frac {5}{3}\sqrt {3},5\right) \qquad \mathrm{(E)}\ \left(\frac {10}{3}\sqrt {3},5\right)$

Solution

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 $\angle AOB=30^\circ$.

By the Pythagorean theorem we have $AO^2 = AB^2 + BO^2 = AB^2 + 25$.

By the definition of sine, we have $\frac{AB}{AO} = \sin AOB = \sin 30^\circ = \frac 12$, hence $AO=2\cdot AB$.

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 $\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }$.

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AMC 10 Problems and Solutions

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