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Difference between revisions of "2008 AMC 10B Problems/Problem 14"

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{{empty}}
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==Problem==
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Triangle <math>OAB</math> has <math>O=(0,0)</math>, <math>B=(5,0)</math>, and <math>A</math> in the first quadrant. In addition, <math>\angle ABO=90^\circ</math> and <math>\angle AOB=30^\circ</math>. Suppose that <math>OA</math> is rotated <math>90^\circ</math> counterclockwise about <math>O</math>. What are the coordinates of the image of <math>A</math>?
  
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==Problem==
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<math>
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{{problem}}
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\mathrm{(A)}\ \left( - \frac {10}{3}\sqrt {3},5\right)
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\qquad
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\mathrm{(B)}\ \left( - \frac {5}{3}\sqrt {3},5\right)
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\qquad
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\mathrm{(C)}\ \left(\sqrt {3},5\right)
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\qquad
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\mathrm{(D)}\ \left(\frac {5}{3}\sqrt {3},5\right)
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\qquad
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\mathrm{(E)}\ \left(\frac {10}{3}\sqrt {3},5\right)
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</math>
  
 
==Solution==
 
==Solution==
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As <math>\angle ABO=90^\circ</math> and <math>A</math> in the first quadrant, we know that the <math>x</math> coordinate of <math>A</math> is <math>5</math>. We now need to pick a positive <math>y</math> coordinate for <math>A</math> so that we'll have <math>\angle AOB=30^\circ</math>.
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By the [[Pythagorean theorem]] we have <math>AO^2 = AB^2 + BO^2 = AB^2 + 25</math>.
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By the definition of [[sine]], we have <math>\frac{AB}{AO} = \sin AOB = \sin 30^\circ = \frac 12</math>, hence <math>AO=2\cdot AB</math>.
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Substituting into the previous equation, we get <math>AB^2 = \frac{25}3</math>, hence <math>AB=\frac{5\sqrt 3}3</math>.
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This means that the coordinates of <math>A</math> are <math>\left(5,\frac{5\sqrt 3}3\right)</math>.
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After we rotate <math>A</math> <math>90^\circ</math> counterclockwise about <math>O</math>, it will get into the second quadrant and have the coordinates
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<math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }</math>.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2008|ab=B|num-b=13|num-a=15}}

Revision as of 15:48, 11 February 2009

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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