Cap. 4 - Problemas Adicionais ⇒ Problema 107 - Relaciones Métricas -Vol. 8 Tópico resolvido

[petrasMOD]

Na figura mostrada se AB = 13m; BC= 14m; AC = 15m e m[tex3]\overset{\LARGE{\frown}}{BM}[/tex3]=m[tex3]\overset{\LARGE{\frown}}{MC}[/tex3].
Calcular "x".
A) 9m
B) 13,5m
C) [tex3]\sqrt{107}[/tex3]m
D) [tex3]\sqrt{111}[/tex3]m
E) [tex3]\sqrt{113}[/tex3]m

Resposta

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Resposta:E

[petrasMOD]

O = centro do circuncentro de △ABC
h = altura do triângulo (isósceles) △OBM
[tex3]{\mathsf{
cos(PBM)=cos(\angle ABC+ \angle CBM)= cos(\angle ABC)cos(\angle CBM)−sen(\angle ABC)sen(\angle CBM)\\
BM=MC \implies \angle CBM=\frac{\angle CAB}{2}=\frac{\angle OBM}{2}\\
cos(\frac{\angle OBM}{2})=\frac{h}{OB}=\frac{h}{R}\\
sen(\frac{\angle OBM}{2})=\frac{BM}{2OB}=\frac{BM}{2R}\\
T,cossenos △ABC\\
15^2=13^2+14^2−2⋅13⋅14cos(\angle ABC)⟹cos(\angle ABC)=\frac{5}{13}\\
sen(\angle ABC)=\sqrt{1−\frac{25}{169}}=\frac{12}{13}\\
cos(\angle PBM)=\frac{5h}{13R}−\frac{12BM}{26R}\\
R=\frac{AC}{2sen(\angle ABC)}=\frac{15⋅13}{24}=\frac{6}{58}\\
BM^2=2R^2−2R^2cos(\angle CAB)=2R^2(1−\frac{33}{65})=65⟹BM=\sqrt{65}\\
h^2=R^2−\frac{BM^2}{4}⟹h=\frac{7\sqrt{65}}{8}\\
Sustituindo:cos(\angle PBM)=\frac{5⋅7\sqrt{65}}{13⋅65}−\frac{8⋅12\sqrt{65}}{26⋅65}=−\frac{1}{\sqrt{65}}\\
PB=13+14+152−15=6,\\
T.cosseno △PBM:\\
PM^2=6^2+65+2⋅6⋅\frac{\sqrt{65}}{\sqrt{65}}=113⟹PM=\sqrt{113}
}}[/tex3]
(Solução:Pie)