Triângulos - 2008 - Vol. 2 ⇒ 015 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Da figura, calcular "x" , se m[tex3]\angle\frac{PMA}{2}[/tex3] = m [tex3]\angle \frac{PMQ}{3} [/tex3]=m [tex3]\frac{QMB }{4}[/tex3]

Resposta

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Gabarito: 30^o

[petrasMOD]

[tex3]\mathsf{
\angle PMA = a, \angle PMQ = b, \angle QMB = c:\\
\frac{a}{2} = \frac{b}{3} = \frac{c}{4} \implies b=\frac{3a}{2}:c = 2a\\
a+b+c = 180^o \implies a+\frac{3a}{2}+2a = 180 \therefore a = 40^o:b = 60^o :c = 80^o \\
\triangle APM: \alpha +\beta = 140^o(I)\\
\triangle PMQ: \alpha+ \phi = 120^o(II) \\
\triangle MQB: \theta + \phi = 100^o(III) \\
(I) +(III): \underbrace{\alpha+\phi}_{120}+\theta +\beta = 240 \implies \theta +\beta = 120^o \\
180-x+90+180-2\beta+180-2\theta =360 \implies x+2(\underbrace{\beta +\theta}_{120})=270
\therefore \boxed{x =30^ o}
}[/tex3]