Question

Answer 1

\(\Delta ABC\) vuông tại A có:

\(cos\widehat{ABC}=cos60^0=\dfrac{AB}{BC}=\dfrac{1}{2}\Rightarrow AB=\dfrac{BC}{2}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\left(cm\right)\)

\(sin\widehat{ABC}=sin60^0=\dfrac{AC}{BC}=\dfrac{\sqrt{3}}{2}\Rightarrow AC=\dfrac{\sqrt{3}}{2}BC=\dfrac{\sqrt{3}}{2}.2\sqrt{3}=3\left(cm\right)\)

\(AH.BC=AB.AC\) (hệ thức lượng trong tam giác vuông)

\(\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{\sqrt{3}.3}{2\sqrt{3}}=\dfrac{3}{2}\left(cm\right)\)