Question

Determine em radianos( exercício na imagem)​

Answer 1

$\large\boxed{\begin{array}{l}\sf complementar~do~\hat angulo\,x\,qualquer:\dfrac{\pi}{2}-x\\\\\sf suplementar\,do\,\hat angulo\,x\,qualquer:\pi-x\end{array}}$

$\large\boxed{\begin{array}{l}\rm a)~\sf\dfrac{\pi}{2}-\dfrac{5\pi}{12}=\dfrac{6\pi-5\pi}{12}=\dfrac{\pi}{12}\\\\\rm b)~\sf\pi-\dfrac{7\pi}{10}=\dfrac{10\pi-7\pi}{10}=\dfrac{3\pi}{10}\end{array}}$