$BC=HB+HC=9+16=25cm$
Xét $ΔABC$ vuông tại $A$
$AM$ là đường trung tuyến ứng cạnh huyền $BC$
$→AM=\dfrac{BC}{2}=\dfrac{25}{2}=12,5cm$
Áp dụng định lý Pytago vào $ΔAHM$ vuông tại $H$
$→HM=\sqrt{AM^2-AH^2}=\sqrt{12,5^2-12}=\sqrt{156,25-144}=\sqrt{12,25}=3,5cm$
Xét $ΔAHM$:
$\cos\widehat{HAM}=\dfrac{AH}{AM}=\dfrac{12}{12,5}=\dfrac{24}{25}$
$\sin\widehat{HAM}=\dfrac{HM}{AM}=\dfrac{3,5}{12,5}=\dfrac{7}{25}$
$\tan\widehat{HAM}=\dfrac{HM}{AH}=\dfrac{3,5}{12}=\dfrac{7}{24}$
$\cot\widehat{HAM}=\dfrac{AH}{HM}=\dfrac{12}{3,5}=\dfrac{24}{7}$
Vậy $\cos\widehat{HAM}=\dfrac{24}{25},\sin\widehat{HAM}=\dfrac{7}{25},\tan\widehat{HAM}=\dfrac{7}{24},\cot\widehat{HAM}=\dfrac{24}{7}$
