Question

calcule o sen ≈ e tg ≈ , sabendo que o cos = ≈ = 12/13

Answer 1

$\text{sen}^2~x+\text{cos}^2~x=1$

$\text{sen}^2~x+\left(\dfrac{12}{13}\right)^2=1~~\Rightarrow~~\text{sen}^2~x+\dfrac{144}{169}=1~~\Rightarrow~~\text{sen}^2~x=1-\dfrac{144}{169}$

$\text{sen}^2~x=\dfrac{25}{169}~~\Rightarrow~~\text{sen}~x=\sqrt{\dfrac{25}{169}}~~\Rightarrow~~\boxed{\text{sen}~x=\dfrac{5}{13}}$

$\text{tg}~x=\dfrac{\text{sen}~x}{\text{cos}~x}$

$\text{tg}~x=\dfrac{\dfrac{5}{13}}{\dfrac{12}{13}}~~\Rightarrow~~\text{tg}~x=\dfrac{5}{13}\cdot\dfrac{13}{12}~~\Rightarrow~~\boxed{\text{tg}~x=\dfrac{5}{12}}$