a: Xét ΔBAC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{a^2+AC^2-7a^2}{2\cdot a\cdot AC}=\dfrac{-1}{2}\)
=>\(2\left(AC^2-6a^2\right)=-2a\cdot AC\)
=>\(AC^2-6a^2=AC\cdot-a\)
=>\(AC^2+AC\cdot a-6a^2=0\)
=>AC^2+3*AC*a-2*AC*a-6a^2=0
=>AC(AC+3a)-2a(AC+3a)=0
=>AC=2a
Xét ΔBAC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}=\dfrac{a^2+7a^2-4a^2}{2\cdot a\cdot a\sqrt{7}}=\dfrac{2\sqrt{7}}{7}\)
nên góc B=41 độ
=>góc C=180-120-41=60-41=19 độ
b: \(m_A=\sqrt{\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}}=\sqrt{\dfrac{a^2+4a^2}{2}-\dfrac{7a^2}{4}}=\dfrac{\sqrt{3}}{2}\cdot a\)
\(\dfrac{BC}{sinA}=2\cdot R\)
=>\(2\cdot R=\dfrac{a\sqrt{7}}{sin120}=a\sqrt{7}\cdot\dfrac{2}{\sqrt{3}}\)
=>\(R=a\sqrt{\dfrac{7}{3}}\)
