Kẻ CH vuông góc AB
\(\Rightarrow CH=BC.sin50\) ; \(CH=AC.sin30\Rightarrow BC.sin50=AC.sin30\)
\(\Rightarrow BC=\frac{AC.sin30}{sin50}\Rightarrow AC+BC=\frac{AC\left(sin30+sin50\right)}{sin50}\)
\(\Rightarrow AC\left(AC+BC\right)=\frac{AC^2\left(sin50+sin30\right)}{sin50}\)
\(BH=BC.cos50\) ; \(AH=AC.cos30\)
\(\Rightarrow BC.cos50+AC.cos30=AB=2\sqrt{3}\)
\(\Rightarrow\frac{AC.sin30}{sin50}.cos50+AC.cos30=2\sqrt{3}\)
\(\Rightarrow AC\left(\frac{sin30.cos50+cos30.sin50}{sin50}\right)=2\sqrt{3}\)
\(\Rightarrow AC=\frac{2\sqrt{3}.sin50}{sin30.cos50+cos30.sin50}\)
\(\Rightarrow AC\left(AC+BC\right)=\frac{12.sin^250}{\left(sin30.cos50+cos30.sin50\right)}.\frac{\left(sin30+sin50\right)}{sin50}=...\)
@Akai Haruma có cách khác không ạ, cách của anh Lâm mình tính số hơi mệt
