Problem 363: (2010 AMC 12b, 13)
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
- Since $\cos(2A-B)\le1$ and $\sin(A+B)\le1$, equality forces $\cos(2A-B)=1$ and $\sin(A+B)=1$: $B=2A$ and $A+B=90^\circ$.
- Solving: $3A=90^\circ$, so $A=30^\circ,B=60^\circ,C=90^\circ$.
- $BC=AB\sin A=4\cdot\frac12=2$, answer (C).
In $\triangle ABC$, $\sin(2A-B)+\sin(A+B)=2$ and $AB=6$. What is $BC$?
