Solucionário:Racso - Cap III - Problemas de Geometria y Como Resolverlos - I Edição - Ex:29
Enviado: 20 Jan 2022, 16:32
Problema Proposto
29 - Calcular x, se L1 || L2
E) 141o
29 - Calcular x, se L1 || L2
Resposta
E) 141o
[tex3]\mathsf{\angle MVF = 63^o\\\angle KFJ =90-2\alpha-27 = 63^o-2\alpha\\
\angle KJF = 27+2\alpha \implies 4\alpha+3\alpha +27+2\alpha = 180^o\\
\therefore \alpha = 17^0\implies \angle KJF = 61^o,3\alpha = 51^o ~e ~4\alpha = 68^o\\
2\theta = \angle KJF(alt.int) \therefore \theta = \frac{61^o}{2}\\
\triangle WFV :\angle NWJ = 180-90-63 =27^o \implies \angle WNJ = 85^o\\
\therefore \beta = \frac{85}{2}\\
\triangle PJL:\angle LPJ = 180-3\alpha-\theta = \frac{197^o}{2}\\
\therefore x = \frac{197}{2}+\frac{85}{2} = \boxed{\color{red}{141}^o}
}[/tex3]