Cho hai vectơ 

D

- Hướng dẫn giải

$\left|\overrightarrow{a}\right|=\sqrt{1^2+{\left(\sqrt{3}\right)}^2}=2, \left|\overrightarrow{b}\right|=\sqrt{{\left(-2\sqrt{3}\right)}^2+{\left(6\right)}^2}=\sqrt{48}=4\sqrt{3}$

$\overrightarrow{a}.\overrightarrow{b}=1.\left(-2\sqrt{3}\right)+\sqrt{3}.6=4\sqrt{3}$

$\cos \left(\overrightarrow{a}, \overrightarrow{b}\right)=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{4\sqrt{3}}{2.4\sqrt{3}}=\frac{1}{2}\Rightarrow \left(\overrightarrow{a},\overrightarrow{b}\right)=60$

Chọn D