smc

Let $\angle A=x^{\circ }$, so $\angle C=3x^{\circ }$, and thus $\angle B={\left(180-4x\right)}^{\circ }$. Now let $D$ be a point on side $AB$ such that $\angle ACD=x^{\circ }$, so $\angle BCD=3x^{\circ }-x^{\circ }=2x^{\circ }$, which gives $$\angle CDB=180^{\circ }-2x^{\circ }-{\left(180-4x\right)}^{\circ }=2x^{\circ },$$ meaning that $△CDB$ and $△CDA$ are both isosceles, with $BC=BD$ and $AD=CD$. In particular, $BD=BC=27$ and $CD=AD=AB-BD=48-27=21$. Hence by Stewart's theorem on triangle $ABC$, $$\begin{array}{rl}& BD\cdot AB\cdot AD+C{D}^2\cdot AB=A{C}^2\cdot BD+B{C}^2\cdot AD\\ & ⟺27\cdot 48\cdot 21+21^2\cdot 48=A{C}^2\cdot 27+27^2\cdot 21\\ & ⟺A{C}^2=\frac{27(21)(48-27)+21^2\cdot 48}{27}\\ & ⟺A{C}^2=\frac{21^2(27+48)}{27}\\ & ⟺A{C}^2=\frac{21^2\cdot 25}{9}\\ & ⟺AC=\frac{21\cdot 5}{3}\text{(as }AC>0\text{)}\\ & ⟺AC=\boxed{35}.\end{array}$$