In triangle \(ABC\), \(BC=6\) and \(6\sin A+5\cos B\cos C=9\). Find the area of \(ABC\).
Solution
Answer: \(6\)
Using the product-to-sum identity and \(B+C=\pi-A\),
\[\cos B\cos C=\frac{\cos(B+C)+\cos(B-C)}2=\frac{-\cos A+\cos(B-C)}2.\]
The given equation becomes \(6\sin A-\frac52\cos A+\frac52\cos(B-C)=9\). The first two terms are at most \(\sqrt{6^2+(5/2)^2}=\frac{13}2\), and the last term is at most \(\frac52\). Equality is necessary in both bounds.
Thus \(B=C\), \(\sin A=\frac{12}{13}\), and \(\cos A=-\frac5{13}\). It follows that \(AB=AC=\sqrt{13}\), so the area is \(\frac12\cdot13\cdot\frac{12}{13}=6\).