Question

3
Sabe-se que sen x = 2/3 e que x é um ângulo obtuso. Determine cos x e tg x.
(1 Ponto)​

Answer 1

$\mathsf{sen(x)=\dfrac{2}{3}\to~sen^2(x)=\dfrac{4}{9}}\\\mathsf{cos^2(x)=\dfrac{9}{9}-\dfrac{4}{9}=\dfrac{5}{9}}$

$\mathsf{cos(x)=-\sqrt{\dfrac{5}{9}}=-\dfrac{\sqrt{5}}{3}}$

$\mathsf{tg(x)=\dfrac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}=-\dfrac{2\sqrt{5}}{5}}$