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Difference between revisions of "2010 AMC 12B Problems/Problem 13"

(Created page with 'We start by noticing that the maximum values for both the sine and the cosine function are 1. Therefore, the only way for this equation to be true is if <math>\cos(2A-B)=1</math…')
 
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We start by noticing that the maximum values for both the sine and the cosine function are 1.  
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== Problem ==
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In <math>\triangle ABC</math>, <math>\cos(2A-B)+\sin(A+B)=2</math> and <math>AB=4</math>. What is <math>BC</math>?
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<math>\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}</math>
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== Solution ==
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We notice that the maximum values for the sine and the cosine function are both 1.  
 
Therefore, the only way for this equation to be true is if <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if one of these equaled less than 1, the other one would have to be greater than 1 which contradicts our previous statement.
 
Therefore, the only way for this equation to be true is if <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if one of these equaled less than 1, the other one would have to be greater than 1 which contradicts our previous statement.
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From this we easily conclude that <math>2A-B=0</math> and <math>A+B=90</math>. Solving this system gives us <math>A=30</math> and <math>B=60</math>, which gives us a <math>30-60-90</math> triangle. By drawing a diagram we can easily see that <math>BC=2</math>　<math>(C)</math>
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From this we easily conclude that <math>2A-B=0</math> and <math>A+B=90</math> and solving this system gives us <math>A=30</math> and <math>B=60</math>. We can easily see that <math>\triangle ABC</math> is a <math>30-60-90</math> triangle with <math>AB=4</math>, <math>AC=2\sqrt{2}</math>, and <math>BC=2</math>　<math>(C)</math>

Revision as of 20:51, 6 April 2010

Problem

In $\triangle ABC$, $\cos(2A-B)+\sin(A+B)=2$ and $AB=4$. What is $BC$?

$\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}$

Solution

We notice that the maximum values for the sine and the cosine function are both 1. Therefore, the only way for this equation to be true is if $\cos(2A-B)=1$ and $\sin(A+B)=1$, since if one of these equaled less than 1, the other one would have to be greater than 1 which contradicts our previous statement. From this we easily conclude that $2A-B=0$ and $A+B=90$ and solving this system gives us $A=30$ and $B=60$. We can easily see that $\triangle ABC$ is a $30-60-90$ triangle with $AB=4$, $AC=2\sqrt{2}$, and $BC=2$　$(C)$

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