Question

Dado cos α=√5/5,encontre sen α e tg α.

Answer 1

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

$\text{sen}^2~\alpha+\left(\dfrac{\sqrt{5}}{5}\right)^2=1~~\Rightarrow~~\text{sen}^2~\alpha+\dfrac{5}{25}=1~~\Rightarrow~~\text{sen}^2~\alpha=1-\dfrac{5}{25}$

$\text{sen}^2~\alpha=\dfrac{25-5}{25}~~\Rightarrow~~\text{sen}^2~\alpha=\dfrac{20}{25}~~\Rightarrow~~\text{sen}~\alpha=\sqrt{\dfrac{20}{25}}$

$\text{sen}~\alpha=\dfrac{2\sqrt{5}}{5}$

$\tex{tg}~\alpha=\dfrac{\text{sen}~\alpha}{\text{cos}~\alpha}$

$\text{tg}~\alpha=\dfrac{\dfrac{2\sqrt{5}}{5}}{\dfrac{\sqrt{5}}{5}}~~\Rightarrow~~\text{tg}~\alpha=2$.