Question

Considere x um ângulo agudo, tal que $sen.x = \frac{2}{5}$ . Calcule :
A) cos x
B) tg x
C) sec x
D) cossec x
E) cotg x

Answer 1

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Sendo um ângulo agudo $\mathsf{x}$ e sabemos que


$\mathsf{sen\,x=\dfrac{\,2\,}{5}\qquad\quad(i)}\\\\ \mathsf{5\,sen\,x=2}$


A) Elevando os dois lados ao quadrado, obtemos

$\mathsf{(5\,sen\,x)^2=2^2}\\\\ \mathsf{25\,sen^2\,x=4}\qquad\textsf{(mas }\mathsf{sen^2 x=1-cos^2\,x}\textsf{)}\\\\ \mathsf{25\cdot (1-cos^2\,x)=4}\\\\ \mathsf{25-25\,cos^2\,x=4}\\\\ \mathsf{25-4=25\,cos^2\,x}$

$\mathsf{25\,cos^2\,x=21}\\\\ \mathsf{cos^2\,x=\dfrac{21}{25}}\\\\\\ \mathsf{cos\,x=\pm\,\sqrt{\dfrac{21}{25}}}\\\\\\ \mathsf{cos\,x=\pm\,\dfrac{\sqrt{21}}{5}}$


Como $\mathsf{x}$ é agudo, o seu cosseno é positivo. Portanto,

$\boxed{\begin{array}{c}\mathsf{cos\,x=\dfrac{\sqrt{21}}{5}} \end{array}}\qquad\quad\checkmark$


B) $\mathsf{tg\,x=\dfrac{sen\,x}{cos\,x}}$

$\mathsf{tg\,x=\dfrac{\;\frac{\sqrt{21}}{5}\;}{\frac{2}{5}}}\\\\\\ \mathsf{tg\,x=\dfrac{\sqrt{21}}{\diagup\!\!\!\! 5}\cdot \dfrac{\diagup\!\!\!\! 5}{2}}\\\\\\ \boxed{\begin{array}{c}\mathsf{tg\,x=\dfrac{\sqrt{21}}{2}} \end{array}}\qquad\quad\checkmark$


C) $\mathsf{sec\,x=\dfrac{1}{cos\,x}}$

$\mathsf{sec\,x=\dfrac{1}{\;\frac{\sqrt{21}}{5}\;}}\\\\\\ \boxed{\begin{array}{c}\mathsf{sec\,x=\dfrac{5}{\sqrt{21}}} \end{array}}\qquad\quad\checkmark$


D) $\mathsf{cossec\,x=\dfrac{1}{sen\,x}}$

$\mathsf{cossec\,x=\dfrac{1}{\;\frac{2}{5}\;}}\\\\\\ \boxed{\begin{array}{c}\mathsf{cossec\,x=\dfrac{\,5\,}{2}} \end{array}}\qquad\quad\checkmark$


E) $\mathsf{cotg\,x=\dfrac{cos\,x}{sen\,x}}$

$\mathsf{cotg\,x=\dfrac{\;\frac{\sqrt{21}}{5}}{\frac{2}{5}}}\\\\\\ \mathsf{cotg\,x=\dfrac{\sqrt{21}}{\diagup\!\!\!\! 5}\cdot \dfrac{\diagup\!\!\!\! 5}{2}}\\\\\\ \boxed{\begin{array}{c}\mathsf{cotg\,x=\dfrac{\sqrt{21}}{2}} \end{array}}\qquad\quad\checkmark$


Bons estudos! :-)