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

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== Solution ==
 
== Solution ==
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We notice that the maximum values for both the sine and the cosine function are 1.  
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We note 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> 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.
 
From this we can easily conclude that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. From this we see that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle with <math>AB=4</math>, <math>AC=2\sqrt{2}</math>, and <math>BC=2</math>　<math>(C)</math>
 
From this we can easily conclude that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. From this we see that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle with <math>AB=4</math>, <math>AC=2\sqrt{2}</math>, and <math>BC=2</math>　<math>(C)</math>

Revision as of 20:01, 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 note 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 $\cos(2A-B)=1$ 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 can easily conclude that $2A-B=0^{\circ}$ and $A+B=90^{\circ}$ and solving this system gives us $A=30^{\circ}$ and $B=60^{\circ}$. From this we see that $\triangle ABC$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with $AB=4$, $AC=2\sqrt{2}$, and $BC=2$　$(C)$

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