Question

Alguém para ajudar?

Answer 1

$\cos \alpha = 0,2\\ \\ \cos \alpha=\dfrac{1}{5}\\ \\ \\ \mathrm{sen^{2\,}}\alpha+\cos^{2}\alpha=1\\ \\ \mathrm{sen^{2\,}}\alpha=1-\cos^{2}\alpha\\ \\ \mathrm{sen^{2\,}}\alpha=1-\left(\dfrac{1}{5} \right )^{2}\\ \\ \mathrm{sen^{2\,}}\alpha=1-\dfrac{1}{25}\\ \\ \mathrm{sen^{2\,}}\alpha=\dfrac{25-1}{25}\\ \\ \mathrm{sen^{2\,}}\alpha=\dfrac{24}{25}\\ \\ \mathrm{sen\,}\alpha=\pm \sqrt{\dfrac{24}{25}}\\ \\ \mathrm{sen\,}\alpha=\pm \dfrac{\sqrt{24}}{\sqrt{25}}\\ \\ \mathrm{sen\,}\alpha=\pm \dfrac{\sqrt{2^{2}\cdot 2 \cdot 3}}{5}\\ \\ \mathrm{sen\,}\alpha=\pm \dfrac{\sqrt{2^{2}}\cdot \sqrt{2 \cdot 3}}{5}\\ \\ \mathrm{sen\,}\alpha=\pm \dfrac{2\sqrt{6}}{5}$


Se $\dfrac{3\pi}{2}<\alpha<2\pi$, então $\alpha$ é do 4º quadrante, e seu seno é negativo. Logo,

$\mathrm{sen\,}\alpha=-\dfrac{2\sqrt{6}}{5}$


Então,

$\mathrm{tg\,}\alpha=\dfrac{\mathrm{sen\,}\alpha}{\cos \alpha}\\ \\ \mathrm{tg\,}\alpha=\dfrac{\,^{-2\sqrt{6}}\!\!\!\diagup\!\!_{5}}{\,^{1}\!\!\!\diagup\!\!_{5}}\\ \\ \mathrm{tg\,}\alpha=-\dfrac{2\sqrt{6}}{5}\cdot 5\\ \\ \mathrm{tg\,}\alpha=-2\sqrt{6}$