Matemática, perguntado por junesangelajunia, 7 meses atrás

Qual deve ser o valor do seno de um ângulo sabendo que Ele se encontra no primeiro quadrante e que o cosseno desse mesmo ângulo é 3/5

Soluções para a tarefa

Respondido por antoniosbarroso2011

Resposta:

Explicação passo-a-passo:

A relação fundamental da trigonometria diz que

sen²x + cos²x = 1

sen² + (3/5)² = 1

sen²x + 9/25 = 1

sen² = 1 - 9/25

sen²x = (25 - 9)/25

sen²x = 16/25

sen x = ± √16/25

sen x = ± 4/5

Como o ângulo é do primeiro quadrante, logo sen x = 4/5

Respondido por procentaury

O seno do ângulo α é 4/5.

Considere o ângulo α do primeiro quadrante na figura a seguir (ou anexa) e as razões trigonométricas do triângulo retângulo, seno e cosseno.

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)(0,0) \thicklines % \thicklines ou \thinlines \put(0,0){\line(1,0){3}} \put(0,0){\line(3,4){3}} \put(3,0){\line(0,1){4}} \qbezier [30](0.6,0.8)(1,0.5)(1,0) \put(2.6,0.4){\line(1,0){0.4}} \put(2.6,0){\line(0,1){0.4}} \put(2.8,0.2){\circle*{0.07}} \put(1.1,0.5){$\alpha$} \put(1.4,-0.5){\sf CA (3)} \put(3.3,1.5){\sf CO} \put(.4,2.3){\large \sf {hip (5)}}\end{picture}$ $\setlength{\unitlength}{1cm} \begin{picture}(6,5)(0,0) \thicklines % \thicklines ou \thinlines\put(0,0){\line(1,0){3}}\put(0,0){\line(3,4){3}}\put(3,0){\line(0,1){4}}\qbezier [30](0.6,0.8)(1,0.5)(1,0)\put(2.6,0.4){\line(1,0){0.4}}\put(2.6,0){\line(0,1){0.4}} \put(2.8,0.2){\circle*{0.07}}\put(1.1,0.5){$\alpha$}\put(1.4,-0.5){\sf CA (3)}\put(3.3,1.5){\sf CO}\put(.4,2.3){\large \sf {hip (5)}}\end{picture}$ $\setlength{\unitlength}{.95cm}\begin{picture}(6,5)(0,0) \thicklines % \thicklines ou \thinlines \put(0,0){\line(1,0){3}} \put(0,0){\line(3,4){3}} \put(3,0){\line(0,1){4}} \qbezier [30](0.6,0.8)(1,0.5)(1,0) \put(2.6,0.4){\line(1,0){0.4}} \put(2.6,0){\line(0,1){0.4}} \put(2.8,0.2){\circle*{0.07}} \put(1.1,0.5){$\alpha$} \put(1.4,-0.5){\sf CA (3)} \put(3.3,1.5){\sf CO} \put(.4,2.3){\large \sf {hip (5)}}\end{picture}$ $\large \text {$ \sf sen\ \alpha = \dfrac{cateto \ oposto}{hipotenusa} $}$

$\large \text {$ \sf cos\ \alpha = \dfrac{cateto \ adjacente }{hipotenusa} $}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)(0,0)\thicklines\put(0,0){\line(1,0){3}}\put(0,0){\line(3,4){3}}\put(3,0){\line(0,1){4}}\qbezier [30](0.6,0.8)(1,0.5)(1,0)\put(2.6,0.4){\line(1,0){0.4}}\put(2.6,0){\line(0,1){0.4}} \put(2.8,0.2){\circle*{0.07}}\put(1.1,0.5){$\alpha$}\put(1.4,-0.5){\sf CA (3)}\put(3.3,1.5){\sf CO}\put(.4,2.3){\large \sf {hip (5)}}\end{picture}$ $\setlength{\unitlength}{1cm}\begin{picture}(6,5)(0,0) \thicklines % \thicklines ou \thinlines \put(0,0){\line(1,0){3}} \put(0,0){\line(3,4){3}} \put(3,0){\line(0,1){4}} \qbezier [30](0.6,0.8)(1,0.4)(1,0) \put(2.6,0.4){\line(1,0){0.4}} \put(2.6,0){\line(0,1){0.4}} \put(2.8,0.2){\circle*{0.07}} \put(1.1,0.5){$\alpha$} \put(1.4,-0.5){\sf CA (3)} \put(3.3,1.5){\sf CO} \put(.4,2.3){\large \sf {hip (5)}}\end{picture}$ $\setlength{\unitlength}{1cm}\begin{picture}(6,5)(0,0) \thicklines % \thicklines ou \thinlines \put(0,0){\line(1,0){3}} \put(0,0){\line(3,4){3}} \put(3,0){\line(0,1){4}} \qbezier [30](0.6,0.8)(1,0.5)(1,0) \put(2.6,0.4){\line(1,0){0.4}} \put(2.6,0){\line(0,1){0.4}} \put(2.8,0.2){\circle*{0.07}} \put(1.1,0.5){$\alpha$} \put(1.4,-0.5){\sf CA (3)} \put(3.3,1.5){\sf CO} \put(.4,2.3){\large \sf {hip (5)}}\end{picture}$

Se o cosseno do ângulo α é 3/5, então o cateto adjacente a α mede 3 unidades e a hipotenusa mede 5 unidades.
Determine a medida do cateto oposto a α usando o teorema de Pitágoras.

hip² = CO² + CA²

5² = CO² + 3²

25 = CO² + 9

CO² = 25 − 9

CO² = 16

CO = 4