2008 AMC 10B Problems/Problem 14

Problem

Triangle [katex]OAB[/katex] has [katex]O=(0,0)[/katex], [katex]B=(5,0)[/katex], and [katex]A[/katex] in the first quadrant. In addition, [katex]\angle ABO=90^\circ[/katex] and [katex]\angle AOB=30^\circ[/katex]. Suppose that [katex]OA[/katex] is rotated [katex]90^\circ[/katex] counterclockwise about [katex]O[/katex]. What are the coordinates of the image of [katex]A[/katex]?

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

Solution 1

Since $\angle ABO=90^\circ$ , and $\angle AOB=30^\circ$ , we know that this triangle is one of the Special Right Triangles.

We also know that $B$ is $(5,0)$ , so $B$ lies on the x-axis. Therefore, $OB = 5$ .

Then, since we know that this is a Special Right Triangle ( $30$ - $60$ - $90$ triangle), we can use the proportion $\frac{5}{\sqrt 3}=\frac{AB}{1}$ to find $AB$ .

We find that $AB=\frac{5\sqrt 3}{3}$

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

Rotate this triangle $90^\circ$ counterclockwise around $O$ , and you will find that $A$ will end up in the second quadrant with the coordinates $\left( -\frac{5\sqrt 3}3, 5\right)$ , or $\boxed{(B)}$

Note: To better visualize this, one can sketch a diagram.

Solution 2

As $\angle ABO=90^\circ$ and $A$ is 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 $OA$ $90^\circ$ counterclockwise about $O$ , it will be in the second quadrant and have the coordinates $\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }$ .

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

Contents

Problem

Solution 1

Solution 2

See also