Áp dụng định lí $Py-ta-go$, ta có:
$BH=\sqrt[]{AB^2-AH^2}=\sqrt[]{3^2-2^2}=\sqrt[]{5}$ $(cm)$
Áp dụng hệ thức lượng, ta có:
$AH^2=BH.CH$
$→ CH=\dfrac{4\sqrt[]{5}}{5}$ $(cm)$
$AC=\sqrt[]{AH^2+HC^2}=\dfrac{6\sqrt[]{5}}{5}$ $(cm)$
$sin\widehat{HAC}=\dfrac{HC}{AC}=\dfrac{2}{3}$
$cos\widehat{HAC}=\dfrac{HA}{AC}=\dfrac{\sqrt[]{5}}{3}$
$tan\widehat{HAC}=\dfrac{HC}{HA}=\dfrac{2\sqrt[]{5}}{5}$
$cot\widehat{HAC}=\dfrac{HA}{HC}=\dfrac{\sqrt[]{5}}{2}$