017 - Triângulos - 2008
Enviado: 18 Ago 2025, 14:00
Do gráfico calcular x, se AB = BC e AD= AC.
Gabarito: 45o
Resposta
Gabarito: 45o
\angle C = 4\beta\\
\triangle ABC: 2\delta = 4\beta+4\beta \implies\delta = 4\beta\\
\triangle ADC_{(isosc.)}: 2\beta+4\beta+4\beta = 180 \implies \beta= 18^o \\
\therefore \delta = 4.18 = 72^o = \angle C \\
\triangle ADC: 2\phi = 2\beta + \angle C = 2.18+ 72 \implies \phi = 54^o \\
\triangle PMD: \theta + \delta = x+\phi \implies x = \theta+72 - 54 = \theta +18(I)\\
\triangle AHD: 180-(x+\theta) +\beta +\phi = 180 \implies x = 72-\theta(II)\\
(I)+(II): 2x = 90 \therefore \boxed{x = 45^o}
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