Cap. 10 - Puntos Notables ⇒ Solucionário:Racso - Cap X - Problemas de Geometria y Como Resolverlos - I Edição - Ex:03 Tópico resolvido

[petrasMOD]

Problema Proposto
3 - No triángulo ABC se sabe que [tex3]\measuredangle EIC-\measuredangle IEC=36º[/tex3].
Sendo "I" e "E", incentro do triânguo ABC e excentro relativo ao lado BC respectivamente; calcular [tex3]\measuredangle ABC.[/tex3].

Resposta

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B) 54^o

[petrasMOD]

[tex3]\measuredangle EIC + \measuredangle IEC +90^o-\theta +\theta = 180^o \implies\boxed{ \measuredangle EIC + \measuredangle IEC = 90^o}\\\begin{cases}\measuredangle EIC + \measuredangle IEC = 90^o \\ \measuredangle EIC -\measuredangle IEC = 36^0\end{cases}\\\therefore \measuredangle EIC = 63^o , ~\measuredangle IEC = 27^o\\\measuredangle B = = 180^o - 2\alpha -2\theta -180^o \rightarrow \boxed{\measuredangle B = 2(\theta - \alpha)}\\\triangle AIC: 63^o = \alpha + 90-\theta \implies \boxed{\theta - \alpha = 27^o} \\
\therefore \boxed{\color{red} \measuredangle ABC = 2.27^o = 54^o}[/tex3]