$cos\widehat{A} = \dfrac{HA}{AB}$
$\Rightarrow HA = AB.cos\widehat{A} = 4.cos60^o = 4.\dfrac{1}{2} = 2 \, (cm)$
Tương tự, $sin\widehat{A} = \dfrac{HB}{AB}$
$\Rightarrow HB = AB.sin\widehat{A} = 4.sin60^o = 4.\dfrac{\sqrt{3}}{2} = 2\sqrt{3} \, (cm)$
$HC = AC - HA = 6 - 2 = 4 \, (cm)$