Question

calcule sen45°, cos45° e tg45° usando a figura abaixo​

Answer 1

$\large\boxed{\begin{array}{l}\rm sen(45^\circ)=\dfrac{\ell}{\ell\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\\\rm cos(45^\circ)=\dfrac{\ell}{\ell\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\\\rm tg(45^\circ)=\dfrac{\ell}{\ell}=1\end{array}}$