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 $A$ ?

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

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 $\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }$ . So the answer is $\boxed{\text{B}}$ .

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2008 AMC 10B Problems/Problem 14

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Solution

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