[tex]BC = 24,\quad \cos(\widehat{B}) = \cos(\widehat{C}) = \dfrac{\sqrt 3}{2}\\ \\ \Rightarrow \widehat{B} = \widehat{C} = 30^\circ,\quad\widehat{A} = 120^\circ\\ \\ \text{Observam ca triunghiul e isoscel} \Rightarrow AB = AC = x\\ \\ \text{Teorema sinusurilor:}\\ \\\dfrac{AB}{\sin (\widehat{C})} = \dfrac{AC}{\sin(\widehat{B})} =\dfrac{BC}{\sin(\widehat{A})} \Rightarrow \dfrac{x}{\sin 30^\circ} = \dfrac{x}{\sin 30^\circ} = \dfrac{24}{\sin 120^\circ} \Rightarrow[/tex]
[tex] \Rightarrow \dfrac{x}{\dfrac{1}{2}} = \dfrac{24}{\dfrac{\sqrt 3}{2}} \Rightarrow 2x = \dfrac{48\sqrt 3}{3}\Rightarrow x = 8\sqrt 3 \Rightarrow\\ \\ \Rightarrow AB = AC = 8\sqrt 3 \\ \\\\ \Rightarrow \boxed{P = 16\sqrt 3+24}[/tex]
