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

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We note that <math>-1</math> <math>\le</math> <math>\sin x</math> <math>\le</math> <math>1</math> and <math>-1</math> <math>\le</math> <math>\cos x</math> <math>\le</math> <math>1</math>.  
 
We note that <math>-1</math> <math>\le</math> <math>\sin x</math> <math>\le</math> <math>1</math> and <math>-1</math> <math>\le</math> <math>\cos x</math> <math>\le</math> <math>1</math>.  
 
Therefore, the only way to satisfy this equation is if both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if either one of these is less than 1, the other one would have to be greater than 1, which contradicts our previous statement.
 
Therefore, the only way to satisfy this equation is if both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if either one of these is 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 can easily conclude that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> (since <math>\cos 0^{\circ}=1</math> and <math>\sin 90^{\circ}=1</math>) and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. It is obvious that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle, and solving for the sides gives us <math>BC=2</math>　<math>\Longrightarrow</math>　<math>(C)</math>
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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>. It is obvious that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle, and solving for the sides gives us <math>BC=2</math>　<math>\Longrightarrow</math>　<math>(C)</math>

Revision as of 22:04, 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 $-1$ $\le$ $\sin x$ $\le$ $1$ and $-1$ $\le$ $\cos x$ $\le$ $1$. Therefore, the only way to satisfy this equation is if both $\cos(2A-B)=1$ and $\sin(A+B)=1$, since if either one of these is 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 $B=60^{\circ}$. It is obvious that $\triangle ABC$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and solving for the sides gives us $BC=2$　$\Longrightarrow$　$(C)$

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