Vol. 09 - Geometria Plana FME 2004 ⇒ 029 - FME 09 -Teste de Vestibulares Tópico resolvido

[petrasMOD]

(FUVEST-91) Na figura, AB = AC, BX = BY e CZ = CY. Se o ângulo A mede 40 °, entâo o ângulo
XYZ mede:

Resposta

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Gabarito: d) 70^o

[petrasMOD]

[tex3]\triangle ABC: \angle B \cong \angle C = \frac{180-40}{2} = 70^o \\
\triangle BXY:\angle
X \cong \angle XYB = \frac{180 - 70}{2} = 55^o \\
\triangle CZY: \angle Z \cong \angle CYZ =\frac{180-70}{2} = 55^o\\
\therefore \angle XYZ = 180 - 55-55 = \boxed{70^o } [/tex3]