Linhas Retas e Ângulos - 2008 - Vol. 1 ⇒ 067 - Retas e Ângulos - 2008 Tópico resolvido

[petrasMOD]

Na figura [tex3]L1 \parallel L2[/tex3]
Calcular PN se PO = 8

Resposta

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Gabarito: 8

[petrasMOD]

[tex3] \mathsf{

\angle CPB =180-90 - (180-2\beta) = 2\beta - 90^o \\
\angle P: 2\beta - 90+2\theta = 90^o \implies \beta+\theta = 90^o \\
\triangle ABP: 2\alpha+\underbrace{\beta +\theta} = 180^o \implies 2\alpha + 90=180 \therefore \alpha= 45^o\\
\angle PON = 180-(\alpha+\beta) = 180 - 45 - \beta = 135^o-\beta\\
\triangle ANP: \angle ANP = 180 - \alpha - \theta -(2\beta-90) = 180 -45+90-(\theta+2\beta) = \\
225-(\underbrace{\theta +\beta}+\beta) = 225-90-\beta \therefore \angle ANP= 135-\beta\\
\therefore \angle ANP = \angle PON \implies \triangle PON_{(isosceles)}: \boxed{PO = PN = 8}
}
[/tex3]