Fórum TutorBrasil - Matemática, Português, Física, Química e Biologia

Problema Proposto
39 - Na figura [tex3]\alpha -\theta = 20^o[/tex3]. Calcular x.

[tex3]\triangle ABC:\measuredangle BCA =90^ o- \alpha \implies \measuredangle DOC = 2\alpha-2x(\triangle DOC~é~isosceles)\\
\overset{\LARGE{\frown}}{ED} = 2x\\ (t.a.externo):\theta = \frac{\overset{\LARGE{\frown}}{CD}- \overset{\LARGE{\frown}}{ED}}{2}=\frac{2(\alpha-\theta)}{2} = \frac{20^o}{2}\\
\therefore \boxed{\color{red}\theta = 10^o} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\triangle ABC:\measuredangle BCA =90^ o- \alpha \implies \measuredangle DOC = 2\alpha-2x(\triangle DOC~é~isosceles)\\
\overset{\LARGE{\frown}}{ED} = 2x\\ (t.a.externo):\theta = \frac{\overset{\LARGE{\frown}}{CD}- \overset{\LARGE{\frown}}{ED}}{2}=\frac{2(\alpha-\theta)}{2} = \frac{20^o}{2}\\
\therefore \boxed{\color{red}\theta = 10^o} [/tex3]