Triângulos - 2008 - Vol. 2 ⇒ 002 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Calcular "x" : se:
m [tex3]\angle[/tex3] ACB - m [tex3]\angle[/tex3] BAC = 96º

Resposta

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Gabarito: 26

[petrasMOD]

[tex3]\mathsf{
\angle A = 180-74 - (180-(x+\theta)) = x+\theta - 74\\
\angle C = 180-A - (180-2\theta) = 254-x-\theta-180+2\theta\\
\angle C = 74-x+\theta\\
C-A = 96 (I)\\
A+C =x+\theta-74+74 - x+\theta =2\theta(II)\\
(II)-(I): 2A = 2\theta -96\ \implies A = \theta-48\\
\triangle: 106= A+180-x+\theta\ = \theta - 48+180-x+\theta = -x+132
\\
\therefore\boxed{x = 26^o}


}[/tex3]