-Giả sử HC:HB=4:3; cạnh AB=14cm.
Có: \(AH^2=HB.HC\Rightarrow\dfrac{AH}{\sqrt{12}}=\dfrac{HB}{3}=\dfrac{HC}{4}\)
Xét \(\Delta_vABH\): \(\tan B=\dfrac{AH}{BH}=\dfrac{\sqrt{12}}{3}\)\(\Rightarrow\widehat{B}\approx49^o\)
\(\Rightarrow AC=AB.\tan B=14.\tan49^o\approx16,1cm\)
\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{14^2+16,1^2}\approx21,34cm\)
Vậy AB=14cm, AC=16,1 cm, BC=21,34 cm.
-Nếu AC=14cm thì AB=16,1 cm; BC=21,34 cm.