Cap. 4 - Triângulos I ⇒ Solucionário:Racso - Cap IV - Problemas de Geometria y Como Resolverlos - I Edição - Ex:11 Tópico resolvido

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Problema Proposto
11 - Na figura achar "x"

Resposta

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C) 60^o

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[tex3]\triangle AEB: \measuredangle EAB + \measuredangle AEB+\measuredangle EBA = 180^o\rightarrow \\180^o -ω-θ+180^o-θ+ 2b+a = 180^o\therefore \boxed{ω+2θ=180^o+a+2b} (I)\\\triangle ADB: \measuredangle DAB + \measuredangle ABD+\measuredangle BDA = 180^o\rightarrow \\180^o -ω-θ+180^o-ω+ 2a+b = 180^o\therefore \boxed{2ω+θ=180^o+2a+b} (II)\\De(I) e(II): ω+2θ-a-2b = 2ω+θ-2a-b \therefore \boxed{a-b = ω-θ}(III)\\ \measuredangle EXC = \measuredangle DXA = 180^\circ-(\theta+b) = 180^\circ-\theta-b\\
\triangle EXC: x=180^\circ-(2b+180^\circ-\theta-b)=\therefore \boxed{x=\theta -b}(IV)\\(IV)e(III): \boxed{\theta -b = ω-a = x}(VII)\\(V)e(II): 2(ω-a)+\theta-\beta = 180^o\rightarrow 2x+x = 180^o \\\therefore \boxed{\color{red}x=60^0}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\triangle AEB: \measuredangle EAB + \measuredangle AEB+\measuredangle EBA = 180^o\rightarrow \\180^o -ω-θ+180^o-θ+ 2b+a = 180^o\therefore \boxed{ω+2θ=180^o+a+2b} (I)\\\triangle ADB: \measuredangle DAB + \measuredangle ABD+\measuredangle BDA = 180^o\rightarrow \\180^o -ω-θ+180^o-ω+ 2a+b = 180^o\therefore \boxed{2ω+θ=180^o+2a+b} (II)\\De(I) e(II): ω+2θ-a-2b = 2ω+θ-2a-b \therefore \boxed{a-b = ω-θ}(III)\\ \measuredangle EXC = \measuredangle DXA = 180^\circ-(\theta+b) = 180^\circ-\theta-b\\
\triangle EXC: x=180^\circ-(2b+180^\circ-\theta-b)=\therefore \boxed{x=\theta -b}(IV)\\(IV)e(III): \boxed{\theta -b = ω-a = x}(VII)\\(V)e(II): 2(ω-a)+\theta-\beta = 180^o\rightarrow 2x+x = 180^o \\\therefore \boxed{\color{red}x=60^0}[/tex3]