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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, there is no other way to satisfy this equation other than making both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since any other way would cause one of these values to become greater than 1, which contradicts our previous statement.
 
Therefore, there is no other way to satisfy this equation other than making both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since any other way would cause one of these values to become 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> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. It is clear that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle with <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 clear that <math>\triangle ABC</math> is a <math>30^{\circ},60^{\circ},90^{\circ}</math> triangle with <math>BC=2</math>　<math>\Longrightarrow</math>　<math>(C)</math>.
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== See also ==
 +
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 00:31, 18 January 2020

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, there is no other way to satisfy this equation other than making both $\cos(2A-B)=1$ and $\sin(A+B)=1$, since any other way would cause one of these values to become 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 clear that $\triangle ABC$ is a $30^{\circ},60^{\circ},90^{\circ}$ triangle with $BC=2$　$\Longrightarrow$　$(C)$.

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 12 Problems and Solutions

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