a, Theo định lí cos
\(BC^2=AB^2+AC^2-AB.AC.2cosA=21\Rightarrow BC=\sqrt{21}\)cm
b, \(cosB=\dfrac{AB^2+BC^2-AC^2}{2AB.BC}=\dfrac{\sqrt{21}}{7}\Rightarrow\widehat{B}\approx49^0\)
-> ^C = 1800 - ^A - ^B = 710
c, Đường trung tuyến xuất phát từ đỉnh A là
\(m_a^2=\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}=\dfrac{61}{4}\Leftrightarrow m_a=\dfrac{\sqrt{61}}{2}\)
ADCT : Nửa chu vi là \(\dfrac{9+\sqrt{21}}{2}\)
\(S=\sqrt{\dfrac{9+\sqrt{21}}{2}\left(\dfrac{9+\sqrt{21}}{2}-5\right)\left(\dfrac{9+\sqrt{21}}{2}-4\right)\left(\dfrac{9+\sqrt{21}}{2}-\sqrt{21}\right)}\)
\(=5\sqrt{3}\left(đvdt\right)\)
\(R=\dfrac{AB.AC.BC}{20\sqrt{3}}=\sqrt{7}\)
\(r=\dfrac{5\sqrt{3}}{3,14}\approx2,75\)
