Vol. 09 - Geometria Plana FME 2004 ⇒ 074 - FME 09 -Teste de Vestibulares Tópico resolvido

[petrasMOD]

(CESGRANRlO-87) Se, na figura, [tex3]\overset{\LARGE{\frown}}{AB} = 20°, \overset{\LARGE{\frown}}{BC} = 124º, \overset{\LARGE{\frown}}{CD} = 36°[/tex3] e [tex3]\overset{\LARGE{\frown}}{DE} = 90°[/tex3], então o ângulo x mede:

Resposta

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Gabarito: c) 37^o

[rcompany]

[tex3]\text{Seja $O$ o centro do círculo, $F$ on ponto de intersecção de $(DE)$ e $(BC)$}\\
\angle EOD=90°\text{ e } OE=OD\implies \angle EDO=45°\\
\angle DOC=36°\text{ e } OD=OC\implies \angle ODC=\angle OCD=\frac{1}{2}\angle DOC=72°\\
\angle COB=124°\text{ e }OB=OC\implies \angle OCB=\angle OBC=\frac{1}{2}(180-\angle COB)=28°\\
\angle FDC=180-\angle EDO-\angle DOC=180-45-72=63°\\
\angle DCF=180- \angle BCO-\angle OCD=180-28-72=80°\\
\angle DFC=180-\angle FDC-\angle DCF=180-63-80=37°[/tex3]