Triângulos - 2008 - Vol. 2 ⇒ 014 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Segundo o gráfico calcular [tex3]\alpha+\beta+\theta[/tex3].
Se NA = AT, MB = BR e CS = CP.

Resposta

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Gabarito: 270^o

[petrasMOD]

[tex3]\mathsf{
\triangle ABC: 2a+2b+2c =180 \implies a+b+c =90^o \\
\triangle RST: 180-\alpha+b+180-\beta+c+180-\theta+a = 180 \implies \underbrace{a+b+c}_{90} - (\alpha+\beta+\theta) = -360 \\
\therefore\boxed{\alpha+\beta+\theta = 270^ o}
}[/tex3]