2010 AMC 12B Problems/Problem 13

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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2010 AMC 12B Problems/Problem 13

Problem

Solution