Question

Em uma circunferência trigonométrica o ângulo de -100° localiza-se no quadrante:

A) 1°
B) 2°
C) 3°
D) 4°​

Answer 1

Resposta:

Opção:    C)

Explicação passo-a-passo:

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.     O ângulo de - 100°  (ângulo negativo)  localiza-se no 3° qua-

.     drante

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(Espero ter colaborado)

Answer 2

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$\Huge\green{\boxed{\rm~~~\red{ C)}~\blue{ 3^{\circ} }~~~}}$

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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, Kayla, 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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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\vector(1,0){5}}\put(0,0){\vector(0,1){5}}\put(0,0){\vector(-1,0){5}}\put(0,0){\vector(0,-1){5}}\put(4.8,0.2){x}\put(0.2,4.8){y}\bezier(-3,0)(-2.77,2.77)(0,3)\bezier(3,0)(2.77,2.77)(0,3)\bezier(-3,0)(-2.77,-2.77)(0,-3)\bezier(3,0)(2.77,-2.77)(0,-3)\put(-1.8,1.5){\Huge\bf{Q_2}}\put(0.7,1.5){\Huge\bf{Q_1}}\put(-1.7,-1.8){\Huge\bf{Q_3}}\put(0.7,-1.8){\Huge\bf{Q_4}}\put(-6,3.5){\dashbox{0.1}(3.4,1){ \sf 90^{\circ} < Q_2 < 180^{\circ} }}\put(3,3.5){\dashbox{0.1}(3.4,1){ \sf 0^{\circ} < Q_1 < 90^{\circ} }}\put(3,-4.2){\dashbox{0.1}(3.4,1){ \sf 270^{\circ} < Q_4 < 360^{\circ} }}\put(-6,-4.2){\dashbox{0.1}(3.4,1){ \sf 180^{\circ} < Q_3 < 270^{\circ} }}\bezier(0.7,0)(0.65,0.65)(0,0.7)\put(0.0,0.67){\line(7,28){0.38}}\bezier(-0.7,0)(-0.65,0.65)(0,0.7)\put(-0.29,0.4){\line(-4,-40){0.38}}\bezier(-0.7,0)(-0.65,-0.65)(0,-0.7)\put(0,-0.3){\line(-4,-22){0.38}}\bezier(0.7,0)(0.65,-0.65)(0,-0.7)\put(1.07,0){\line(-4,-31){0.38}}\put(5.4,-0.2){\LARGE\text{ \mathbf {0^{\circ}} }}\put(-0.4,5.5){\LARGE\text{ \mathbf {90^{\circ}} }}\put(-6.3,-0.1){\LARGE\text{ \mathbf {180^{\circ}} }}\put(-0.5,-5.8){\LARGE\text{ \mathbf {270^{\circ}} }}\end{picture}$

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

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☔ Temos que, sendo o nosso ângulo X negativo, devemos

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1. Extrair todas as voltas completas do ângulo (n * 360º);
2. Encontrar a soma de 360º + x;
3. Encontrar em qual quadrante ele se localiza.

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1)$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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☔ -100º já é um valor que não possui nenhuma volta completa (-100º > -360º);

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2)$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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➡ $\large\blue{\sf 360 + (-100) }$

$\large\blue{\sf = 360 - 100 }$

$\large\blue{\sf = 260 }$

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3)$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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☔ Temos que o ângulo de 260º se encontra no terceiro quadrante, que varia de 180º até 270º

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$\Huge\green{\boxed{\rm~~~\red{ C)}~\blue{ 3^{\circ} }~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\Large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

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

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$