Question

se sen £ = √3/5, então cos £ = e tg £ é .

Answer 1

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De acordo com o texto do enunciado e com a foto em anexo, temos um ângulo agudo $\mathsf{\alpha,}$ de modo que

$\mathsf{sen\,\alpha=\dfrac{\sqrt{3}}{5}}\\\\\\ \mathsf{5\,sen\,\alpha=\sqrt{3}}$


Elevando os dois lados ao quadrados, obtemos

$\mathsf{(5\,sen\,\alpha)^2=(\sqrt{3})^2}\\\\ \mathsf{25\,sen^2\,\alpha=3}\qquad\quad\textsf{(mas }\mathsf{sen^2\,\alpha=1-cos^2\,\alpha}\textsf{)}\\\\ \mathsf{25\cdot (1-cos^2\,\alpha)=3}\\\\ \mathsf{25-25\,cos^2\,\alpha=3}$

$\mathsf{25-3=25\,cos^2\,\alpha}\\\\ \mathsf{22=25\,cos^2\,\alpha}\\\\ \mathsf{cos^2\,\alpha=\dfrac{22}{25}}$

$\mathsf{cos\,\alpha=\pm\,\sqrt{\dfrac{22}{25}}}\\\\\\ \mathsf{cos\,\alpha=\pm\,\dfrac{\sqrt{22}}{5}}$


Mas como $\mathsf{\alpha}$ é agudo, o cosseno é positivo. Portanto,

$\boxed{\begin{array}{c}\mathsf{cos\,\alpha=\dfrac{\sqrt{22}}{5}} \end{array}}\qquad\quad\checkmark$


Calculando a tangente:

$\mathsf{tg\,\alpha=\dfrac{sen\,\alpha}{cos\,\alpha}}\\\\\\ \mathsf{tg\,\alpha=\dfrac{~~\frac{\sqrt{3}}{5}~~}{\frac{\sqrt{22}}{5}}}\\\\\\ \mathsf{tg\,\alpha=\dfrac{\sqrt{3}}{\diagup\!\!\!\! 5}\cdot \dfrac{\diagup\!\!\!\! 5}{\sqrt{22}}}$

$\mathsf{tg\,\alpha=\dfrac{\sqrt{3}}{\sqrt{22}}}\\\\\\ \mathsf{tg\,\alpha=\dfrac{\sqrt{3}\cdot \sqrt{22}}{\sqrt{22}\cdot \sqrt{22}}}\\\\\\ \mathsf{tg\,\alpha=\dfrac{\sqrt{3\cdot 22}}{(\sqrt{22})^2}}\\\\\\\\ \boxed{\begin{array}{c}\mathsf{tg\,\alpha=\dfrac{\sqrt{66}}{22}} \end{array}}\qquad\quad\checkmark$


Bons estudos! :-)


Tags:   razões trigonométricas ângulo agudo seno sen cosseno cos tangente tg tan relação fundamental trigonometria

Answer 2

sen £ = √3/5
cos £ =
 tg £ = 

sen² £ + cos² £ = 1 ==> cos² £ = 1 - (√3/5)² ==> 1 - 3/25 

cos² £ = 25-3 ==> cos² £ = 22 ==> cos £ =  √22 
               25                        25                        5           √



tg £ =  sen £ =   √3/5   =   √3  ==> √3.√22    ==> √66
           cos £       √22/5    √22          √22.√22          22