Question

O valor do ângulo abaixo, é:
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Answer 1

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$\Huge\green{\boxed{\rm~~~\gray{\alpha}~\pink{=}~\blue{ 80^{\circ} }~~~}}$

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$\bf\large\green{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}}$

$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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☺lá, Larissa, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo. ✌

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☔ Inicialmente  vamos relembrar a seguinte propriedade das circunferências:

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• Um ângulo central (<180º) que "vê" um arco sempre terá o dobro do valor do ângulo inscrito que "vê" o mesmo arco, contanto que o ponto que forma o ângulo inscrito não esteja na corda daquele arco.

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☔ Vamos portanto fazer algumas análises geométricas. Inicialmente vamos observar nossos ângulos centrais, que "vêem" os arcos AB e CD

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\bezier(-3,0)(-2.77,2.77)(0,3)\bezier(3,0)(2.77,2.77)(0,3)\bezier(-3,0)(-2.77,-2.77)(0,-3)\bezier(3,0)(2.77,-2.77)(0,-3)\put(-2.86,0.76){\circle*{0.13}}\put(-2.77,-1.06){\circle*{0.13}}\put(2.95,-0.4){\circle*{0.13}}\put(1.84,2.4){\circle*{0.13}}\put(0,0){\circle*{0.13}}\bezier(0,0)(-1.43,0.38)(-2.86,0.77)\bezier(0.0,0.0)(-1.38,-0.53)(-2.77,-1.06)\bezier(0.0,0.0)(1.47,-0.2)(2.95,-0.4)\bezier(0.0,0.0)(0.92,1.2)(1.84,2.4)\bezier(1.1,-0.1)(1,0.7)(0.6,0.8)\bezier(-1.1,-0.4)(-1.4,0)(-1.1,0.3)\put(-2.3,-0.3){\Large \sf 60^{\circ} }\put(1.3,0.6){\Large \sf 100^{\circ} }\put(-0.83,0.38){\circle*{0.13}}\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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☔ Tendo visualizado os ângulos centrais de 60º e 100º vamos agora analisar os ângulos inscritos. Primeiramente vamos observar o ângulo do vértice D que "vê" o arco AB

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\bezier(-3,0)(-2.77,2.77)(0,3)\bezier(3,0)(2.77,2.77)(0,3)\bezier(-3,0)(-2.77,-2.77)(0,-3)\bezier(3,0)(2.77,-2.77)(0,-3)\put(-4,-2){\line(4,3){7}}\put(-2.77,-1.06){\circle*{0.13}}\put(2.95,-0.4){\circle*{0.13}}\put(1.84,2.4){\circle*{0.13}}\bezier{50}(-2.77,-1.06)(0.09,-0.73)(2.95,-0.4)\bezier(-1.6,-0.9)(-1.5,-0.4)(-1.8,-0.35)\put(-1.3,-0.6){\Large \sf 50^{\circ} }\put(-0.83,0.38){\circle*{0.13}}\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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☔ Vamos agora observar o ângulo do vértice B que "vê" o arco DC

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\bezier(-3,0)(-2.77,2.77)(0,3)\bezier(3,0)(2.77,2.77)(0,3)\bezier(-3,0)(-2.77,-2.77)(0,-3)\bezier(3,0)(2.77,-2.77)(0,-3)\put(-4,1){\line(5,-1){8}}\put(-2.86,0.76){\circle*{0.13}}\put(-2.77,-1.06){\circle*{0.13}}\put(2.95,-0.4){\circle*{0.13}}\bezier{50}(-2.77,-1.06)(0.09,-0.73)(2.95,-0.4)\bezier(1.6,-0.55)(1.4,-0.3)(1.6,-0.15)\put(0.6,-0.5){\Large \sf 30^{\circ} }\put(-0.83,0.38){\circle*{0.13}}\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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☔ Com isso temos 2 dos 3 ângulos internos do triângulos DBI.

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\bezier(-3,0)(-2.77,2.77)(0,3)\bezier(3,0)(2.77,2.77)(0,3)\bezier(-3,0)(-2.77,-2.77)(0,-3)\bezier(3,0)(2.77,-2.77)(0,-3)\put(-4,-2){\line(4,3){7}}\put(-4,1){\line(5,-1){8}}\put(-0.83,0.38){\circle*{0.13}}\put(-2.86,0.76){\circle*{0.13}}\put(-2.77,-1.06){\circle*{0.13}}\put(2.95,-0.4){\circle*{0.13}}\put(1.84,2.4){\circle*{0.13}}\bezier(1.6,-0.55)(1.4,-0.3)(1.6,-0.15)\put(0.8,-0.5){\large \sf 30^{\circ} }\bezier{50}(-2.77,-1.06)(0.09,-0.73)(2.95,-0.4)\bezier(-1.6,-0.9)(-1.5,-0.4)(-1.8,-0.35)\put(-1.5,-0.7){\large \sf 50^{\circ} }\bezier(-1.3,0)(-0.7,-0.4)(0,0.2)\put(-0.7,-0.6){\large \sf \beta }\bezier(0,1)(0.6,0.9)(0.6,0.1)\put(1,0.5){\large \sf \alpha }\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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☔ Daqui temos que

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➡ $\large\blue{\sf 30 + 50 + \beta = 180 }$

➡ $\large\blue{\sf 80 + \beta = 180 }$

➡ $\large\blue{\sf \beta = 180 - 80 }$

➡ $\large\blue{\sf \beta = 100 }$

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☔ Analisando os ângulos que nos restam sabemos que α e β são ângulos suplementares, ou seja, juntos formam 180º, ou seja

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➡ $\large\blue{\sf \alpha + \beta = 180 }$

➡ $\large\blue{\sf \alpha + 100 = 180 }$

➡ $\large\blue{\sf \alpha = 180 - 100 }$

➡ $\large\blue{\sf \alpha = 80 }$

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$\Huge\green{\boxed{\rm~~~\gray{\alpha}~\pink{=}~\blue{ 80^{\circ} }~~~}}$ ✅

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_______________________________☁

☕ Bons estudos.

(Dúvidas nos comentários) ☄

__________________________$\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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"Absque sudore et labore nullum opus perfectum est."