Triângulos - 2008 - Vol. 2 ⇒ 018 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Do gráfico calcular a + b

Resposta

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Gabarito: 240^o

[petrasMOD]

[tex3]\mathsf{
\boxed{}_{ABCD}: 180-a+180-b+3\alpha+2\gamma+2\theta+3\theta = 360\\
a+b = 2(\alpha+\beta+\gamma+\theta)\\
\angle BEC = \gamma+3\alpha(I)\\
\angle FEC = \theta+\gamma(II)\\
\therefore (I)+(II) = 180^o \implies 3\alpha+2\gamma+\theta = 180^o (III)\\
\angle EFC = \theta + 2\theta\\
\triangle EFC: \theta +\gamma + \theta +2\beta +\beta =180^o \\
\therefore \gamma+2\theta +3\beta =180^o (IV)\\
(III)+(IV): 3\alpha+3\beta + 3\gamma+3\theta = 360\\
\alpha+\beta +\gamma +\theta = 120^o \\
\therefore a+b = 2(\alpha+\beta +\gamma +\theta) = 2.120 = \boxed{240^o}

}[/tex3]