Theo định lý pytago có : BC^2=AB^2+AC^2
BC^2=6^2+8^2=100
BC=10
AB^2=BH.BC=>BH=AB^2/BC=6^2/10=3,6
AB.AC=AH.BC<=>6.8=AH.10=>AH=6.8/10=4,8
sinBAH=BH/AB=3,6/6=3/5=>cosCAH=3/5
CosBAH=AH/AB=4,8/6=4/5=>SinCAH=4/5
TgBAH=BH/AH=3,6/4,8=3/4=>cotgCAH=3/4
CotgBAH=AH/BH=4,8/3,6=4/3=>tgCAH=4/3