Question

Determine o seno , o cosseno e a tangente dos ângulos a e b​

Answer 1

$\large\boxed{\begin{array}{l}\tt a)~\sf sen(\alpha)=cos(\beta)=\dfrac{1}{2}\\\\\sf cos(\alpha)=sen(\beta)=\dfrac{\sqrt{3}}{2}\\\sf tg(\alpha)=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\\\\\sf tg(\beta)=\dfrac{1}{tg(\alpha)}=\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf sen(\alpha)=cos(\beta)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\\\sf sen(\beta)=cos(\alpha)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\sf tg(\alpha)=\dfrac{1}{1}=1\\\sf tg(\beta)=\dfrac{1}{tg(\alpha)}=\dfrac{1}{1}=1\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf sen(\alpha)=cos(\beta)=\dfrac{3}{5}\\\\\sf cos(\alpha)=sen(\beta)=\dfrac{4}{5}\\\\\sf tg(\alpha)=\dfrac{3}{4}\\\\\sf tg(\beta)=\dfrac{4}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\tt d)~\sf sen(\alpha)=cos(\beta)=\dfrac{2}{2\sqrt{5}}=\dfrac{\backslash\!\!\!2\sqrt{5}}{\backslash\!\!\!2\cdot5}=\dfrac{\sqrt{5}}{5}\\\\\sf cos(\alpha)=sen(\beta)=\dfrac{4}{2\sqrt{5}}=\dfrac{\backslash\!\!\!4\sqrt{5}}{\backslash\!\!\!2\cdot5}=\dfrac{2\sqrt{5}}{5}\\\\\sf tg(\alpha)=\dfrac{2}{4}=\dfrac{1}{2}\\\\\sf tg(\beta)=\dfrac{4}{2}=2\end{array}}$