Question

Determine tg α,dado que sen α= √3/4 e α é angulo agudo.

Answer 1

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

$\left(\dfrac{\sqrt{3}}{4}\right)^2+\text{cos}^2~\alpha=1~~\Rightarrow~~\text{cos}^2~\alpha=1-\dfrac{3}{16}~~\Rightarrow~~\text{cos}^2~\alpha=\dfrac{13}{16}$

$\text{cos}~\alpha=\sqrt{\dfrac{13}{16}}~~\Rightarrow~~\text{cos}~\alpha=\dfrac{\sqrt{13}}{4}$.

$\text{tg}~\alpha=\dfrac{\text{sen}~\alpha}{\text{cos}~\alpha}~~\Rightarrow~~\text{tg}~\alpha=\dfrac{\dfrac{\sqrt{3}}{4}}{\dfrac{\sqrt{13}}{4}}=\dfrac{\sqrt{3}}{\sqrt{13}}~~\Rightarrow~~\text{tg}~\alpha=\dfrac{\sqrt{39}}{13}$;