Question

1) Conhecendo o valor do sen α = 3/5, α ângulo agudo, determine:

a) cos α

b) tg α

2) Dada a cos θ = √5/5, θ ângulo agudo,determine?

a) sen θ

b) tg θ
​

Answer 1

$\Large\boxed{\begin{array}{l}\rm 1)\\\sf sen(\alpha)=\dfrac{3}{5}\implies sen^2(\alpha)=\dfrac{9}{25}\\\tt a)~\sf cos^2(\alpha)=\dfrac{25}{25}-\dfrac{9}{25}=\dfrac{16}{25}\\\\\sf cos(\alpha)=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\\\\\tt b)~\sf tg(\alpha)=\dfrac{\frac{3}{\red{\backslash\!\!\!}5}}{\frac{4}{\red{\backslash\!\!\!}5}}=\dfrac{3}{4}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm 2)\\\sf cos(\theta)=\dfrac{\sqrt{5}}{5}\implies cos^2(\theta)=\dfrac{5}{25}=\dfrac{1}{5}\\\\\tt a)~\sf sen^2(\theta)=\dfrac{5}{5}-\dfrac{1}{5}=\dfrac{4}{5}\\\\\sf sen(\theta)=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\\\\\tt b)~\sf tg(\theta)=\dfrac{\frac{2\sqrt{5}}{\red{\backslash\!\!\!\!}5}}{\frac{1}{\red{\backslash\!\!\!}5}}=2\sqrt{5}\end{array}}$