Theo hệ thức lượng cho tam giác ABC ta có :
\(AB^2=x.BC\Leftrightarrow AB^2=x.\left(x+HC\right)\Leftrightarrow60^2=x\left(x+27\right)\Leftrightarrow x^2+27x-3600=0\)
Ta có : \(x^2+27x-3600=0\)
\(\Delta=27^2+4.3600=15129>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-27+\sqrt{15129}}{2}=48\left(N\right)\\x_2=\dfrac{-27-\sqrt{15129}}{2}=-75\left(L\right)\end{matrix}\right.\)
\(AH^2=x.HC\Rightarrow AH=\sqrt{x.HC}=\sqrt{48.27}=36cm\)
\(AC^2=AH^2+HC^2\Rightarrow AC=\sqrt{AH^2+HC^2}=\sqrt{36^2+27^2}=45cm\)