\(AC+BC=16\) \(\Rightarrow AC=16-BC\)
Theo định lý hàm cos:
\(AB^2=AC^2+BC^2-2AC.BC.cosC\)
\(\Rightarrow AC^2+BC^2+AC.BC=196\)
\(\Rightarrow\left(16-BC\right)^2+BC^2+\left(16-BC\right).BC=196\)
\(\Rightarrow BC^2-16BC+60=0\)
\(\Rightarrow\left[{}\begin{matrix}BC=10\Rightarrow AC=6\\BC=6\Rightarrow AC=10\left(loại\right)\end{matrix}\right.\)


