Ta có \(CH=AC.cos\widehat{C}=35.cos50^o\)
\(AH=AC.sin\widehat{C}=35.sin50^o\)
\(BH=AH.cot\widehat{B}=35.sin50^o.cot60^o\)
\(\Rightarrow BC=BH+CH=35.cos50^o+35.sin50^o.cot60^o\)
\(\Rightarrow S_{ABC}=\frac{AH.BC}{2}=\frac{35.sin50^o\left(35.cos50^o+35.sin50^o.cot60^o\right)}{2}\)