2010 AMC 12B Problems/Problem 13

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We start by noticing that the maximum values for both the sine and the cosine function are 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 $A+B=90$. Solving this system gives us $A=30$ and $B=60$, which gives us a $30-60-90$ triangle. By drawing a diagram we can easily see that $BC=2$ $(C)$