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

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== Solution ==
 
== Solution ==
 
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 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.
 
From this we can easily deduce 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 from this 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>
 
From this we can easily deduce 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 from this 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 22:02, 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 deduce that $2A-B=0^{\circ}$ and $A+B=90^{\circ}$ (since $\cos 0^{\circ}=1$ and $\sin 90^{\circ}=1$) and solving this system gives us $A=30^{\circ}$ and $B=60^{\circ}$. It is obvious from this 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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