2010 AMC 12B Problems/Problem 13: Difference between revisions

Revision as of 09:59, 4 July 2013

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 $BC=2$ 　 $\Longrightarrow$ 　 $(C)$ .

2010 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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2010 AMC 12B Problems/Problem 13: Difference between revisions

Revision as of 09:59, 4 July 2013

Problem

Solution

See also