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 the max value for both sine and cosine of any angle is 1. Therefore, the only way to satisfy the equation is if $\cos(2A-B)=1$ 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 deduce 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 from this that $\triangle ABC$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with $AB=4$ , $AC=2\sqrt{2}$ , and $BC=2$ 　 $(C)$

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2010 AMC 12B Problems/Problem 13

Problem

Solution