a: Sai
b: \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
=>\(\dfrac{1}{2}\cdot6\cdot AC\cdot sin120=12\sqrt{3}\)
=>\(AC\cdot\dfrac{3\sqrt{3}}{2}=12\sqrt{3}\)
=>\(AC\cdot1,5=12\)
=>AC=8
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{6^2+8^2-BC^2}{2\cdot6\cdot8}=cos120=-\dfrac{1}{2}\)
=>\(100-BC^2=2\cdot6\cdot8\cdot\dfrac{-1}{2}=-48\)
=>\(BC^2=148\)
=>\(BC=\sqrt{148}=2\sqrt{37}\)
=>Đúng
c: Xét ΔABC có \(\dfrac{BC}{sinA}=2R\)
=>\(2R=2\sqrt{37}:sin120=2\sqrt{37}:\dfrac{\sqrt{3}}{2}=\dfrac{4\sqrt{37}}{\sqrt{3}}\)
=>\(R=2\sqrt{\dfrac{37}{3}}=\dfrac{2\sqrt{111}}{3}\)
=>Sai
d: Độ dài đường trung tuyến kẻ từ C là:
\(\sqrt{\dfrac{2\left(CA^2+CB^2\right)-AB^2}{4}}=\sqrt{\dfrac{2\left(8^2+148\right)-6^2}{4}}=\sqrt{97}\)
=>Sai