Question

Answer 1

Do góc A tù \(\Rightarrow cosA< 0\Rightarrow cosA=-\sqrt{1-sin^2A}=-\dfrac{4}{5}\)

Áp dụng định lý hàm cos:

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=\sqrt{130}\)

\(S_{ABC}=\dfrac{1}{2}AB.AC.sinA=21\)

\(AM=\sqrt{\dfrac{2\left(AB^2+AC^2\right)-BC^2}{4}}=\dfrac{3\sqrt{2}}{2}\)