\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=\dfrac{2}{3}\)

\(sinA=\sqrt{1-cos^2A}=\dfrac{\sqrt{5}}{3}\)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}AB.AC.sinA=6\sqrt{5}\)

Áp dụng hệ quả định lí cô sin trong tam giác ta có:

  cos A = 2 2 + 2 2 − 1 2 2.2.2 = 7 8

Chọn C.

Xét tam giác ABC nhọn có \(BC^2=AB^2+AC^2-2AB\cdot AC\cdot\cos\widehat{A}\)
\(\Rightarrow\cos\widehat{A}=\dfrac{AB^2+AC^2-BC^2}{2AB\cdot AC}=\dfrac{AB^2+AC^2-BC^2}{4\cdot\dfrac{1}{2}AB\cdot AC}=\dfrac{AB^2+AC^2-BC^2}{4S_{ABC}}\)

Cmtt: \(\left\{{}\begin{matrix}\cos\widehat{B}=\dfrac{AB^2+BC^2-AC^2}{4S_{ABC}}\\\cos\widehat{C}=\dfrac{AC^2+BC^2-AB^2}{4S_{ABC}}\end{matrix}\right.\)
\(\Rightarrow\cos\widehat{A}+\cos\widehat{B}+\cos\widehat{C}\\ =\dfrac{AB^2+AC^2-BC^2+AB^2+BC^2-AC^2+AC^2+BC^2-AB^2}{4S_{ABC}}\\ =\dfrac{AB^2+AC^2+BC62}{4S_{ABC}}\)