a) có tan C =√3
mà tan C=$\frac{AB}{AC}$
⇒$\frac{AB}{AC}$ = √3 = $\frac{k√3}{k}$
t/c : AB²+AC²=BC² (HTL)
⇒BC=√3k²+k²=√4k²=2k
⇒ tanB=$\frac{AC}{AB}$=$\frac{k}{k√3}$=$\frac{1}{√3}$
cotgB=$\frac{AB}{AC}$=$\frac{k√3}{k}$=√3
sinB=$\frac{AC}{BC}$=$\frac{k}{2k}$=$\frac{1}{2}$
cosB=$\frac{AB}{BC}$=$\frac{k√3}{2k}$=$\frac{√3}{2}$
b) t/c: tanC =√3 ⇒ C=60'
t/c: HC= AH.cotC (tslg)
=2√3.$\frac{√3}{3}$
=2
t/c: AH²+HC²=AC² (HTL)
⇒AC=√AH²+HC²=√12+4=√16=4
do tam giác ABC là tam giác vuông nên ∠B =30'
T/c: HB=AH.cotB (tslg)
=2√3.√3
=6
t/c: AB=√AH²+BH² (HTL)
=√12+36
=√48=4√3
T/c: BC=BH+HC=6+2=8
cho mik xin ctlhn nha
