Question

O ângulo α está localizado no Quadrante II, e sen α = 9/41. Qual cos α?

Answer 1

$\mathsf{sen(\alpha)=\dfrac{9}{41}\implies~sen^2(\alpha)=\dfrac{81}{1681}}$ $\mathsf{cos^2(\alpha)=\dfrac{1681-81}{1681}=\dfrac{1600}{1681}}$

$\mathsf{cos(\alpha)=-\sqrt{\dfrac{1600}{1681}}}$

$\large\boxed{\boxed{\boxed{\boxed{\mathsf{cos(\alpha)=-\dfrac{40}{41}}}}}}$