Triângulos - 2008 - Vol. 2 ⇒ 012 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Do gráfico. Calcular "x" sendo o [tex3]\triangle[/tex3]AEB equilátero, e:
[tex3]\alpha+\theta[/tex3] = 140 º

Resposta

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Gabarito: 20^o

[petrasMOD]

[tex3]\mathsf{
\triangle ABC: \angle B = 120+x:\\
120+x+\alpha+\beta = 180^o \implies x = 60^o -\alpha-\beta(I)\\
\triangle ADE: 60-x+\beta +60+180-\theta = 180 \implies x = 120+\beta -\theta(II)\\
(I)+(II) : 2x = 180-(\alpha+\theta) = 180 - 140 = 40 \therefore \boxed{x = 20^o }
}[/tex3]