Resposta
18
Moderador: [ Moderadores TTB ]
VALEU!petras escreveu: ↑22 Abr 2024, 21:57 02rr,
[tex3] \angle APB \cong \angle BPC \cong \angle CPA = \alpha\implies \alpha = 120^0 \\
\angle PAB = \theta \implies \angle PBA = 60-\theta\\
\triangle PAC:\angle PAC = 60^o-\theta \implies \angle ACP = \theta\\
\therefore \triangle ACP \sim\triangle BAP \implies \frac{CP}{AP} =\frac{AB}{BP}\\
\frac{27}{AP} = \frac{AP}{12}\implies AP^2 = 324 \therefore \boxed{AP = 18} [/tex3]