Triângulos - 2008 - Vol. 2 ⇒ 003 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Na figura BM = MN , NQ = QC:
Calcular x

Resposta

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Gabarito:150^o

[petrasMOD]

Usando a propriedade:

[tex3]\mathsf{
x = 90^o + \frac{\angle A}{2}\\
\angle MBN \cong \angle MNB = \alpha\\
\angle QNC \cong \angle QCN = \beta\\
\alpha +\beta +120^o = 180^o \implies \alpha +\beta = 60^o \\
\triangle ABC: \angle A+\alpha+\beta = 180^o \implies \angle A+60 = 180 \therefore \angle A = 120^o \\
\therefore x =90+\frac{120}{2} = \boxed{150^ o}






}[/tex3]