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 notice that the maximum values for the sine and the cosine function are both 1. Therefore, the only way for this equation to be true is if $\cos(2A-B)=1$ and $\sin(A+B)=1$ , since if one of these equaled less than 1, the other one would have to be greater than 1 which contradicts our previous statement. From this we easily conclude that $2A-B=0$ and $A+B=90$ and solving this system gives us $A=30$ and $B=60$ . We can easily see that $\triangle ABC$ is a $30-60-90$ triangle with $AB=4$ , $AC=2\sqrt{2}$ , and $BC=2$ 　 $(C)$

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2010 AMC 12B Problems/Problem 13

Problem

Solution