Xét tg ABC có : A=90 độ; AH \(\perp BC\)

Áp dụng HTL. Ta có: \(AB^2=BH.BC\\ =>BC=\dfrac{AB^2}{BH}=\dfrac{4^2}{3}=\dfrac{16}{3}\)

=> HC= BC- BH=\(\dfrac{16}{3}-3=\dfrac{7}{3}\)

=> \(AC^2=HC.BC=\dfrac{7}{3}.\dfrac{16}{3}=\dfrac{112}{9}\\ =>AC=\dfrac{4\sqrt{7}}{3}\)