Question

Se x pertence ao 1º quadrante e cosx = 4/5 então podemos afirmar que os valores de senx, tgx , cotgx são respectivamente:

a) 3/5 ; 4/3 ; 3/4
b) 1/2 ; 4/3 ; 5/8
c) 1 ; 2/3 ; 3/4
d) 6/11 ; 4/5 ; 5/4
e) 3/5 ; 3/4 ; 4/3

Answer 1

Bom dia Roger!


Solução!


Vamos usar essa relação fundamental da trigonometria para determinarmos o valor do seno.


$sen^{2}(x)+cos^{2}(x)=1$



$cos(x)= \dfrac{4}{5}\\\\\\\ sen^{2}(x)+(\frac{4}{5})^{2} =1\\\\\\\ sen^{2}(x)+\frac{16}{25} =1\\\\\\\ 25sen^{2}(x)+16 =25\\\\\\\ 25sen^{2}(x)=25-16\\\\\\\ 25sen^{2}(x)=9\\\\\\\ sen^{2}(x)= \frac{9}{25}\\\\\\ sen(x)= \sqrt{ \frac{9}{25} } \\\\\\ sen(x)= \frac{3}{5} \\\\\\\ \boxed{Resposta:sen(x)= \frac{3}{5}}$



$Tag(x)= \dfrac{sen(x)}{cos(x)}$


$Sen(x)= \dfrac{3}{5} \\\\\\\ Cos(x)=\dfrac{4}{5}\\\\\\\\\ Tag(x)= \dfrac{3}{ \dfrac{5}{ \dfrac{4}{5} } }\\\\\\\ Tag(x)= \dfrac{3}{5}\times \dfrac{5}{4}\\\\\\\ Tag(x)= \dfrac{3}{4}\\\\\\\ \boxed{Resposta: Tag(x)= \dfrac{3}{4}}$


A cotangente é o inverso da tangente.


$Cotag(x)=tag(x)^{-1}\\\\\\ Cotag(x)= \dfrac{cos(x)}{sen(x)}\\\\\ Cotag(x)= \dfrac{4}{3} \\\\\\\\ \boxed{Resposta:Cotag(x)= \frac{4}{3}}$



$\boxed{Resposta: Alternativa~~E}$


Bom dia!
Bons estudos!