Question

Por que Sen45° é igual a 2 sobre 2?​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Observe\,a\,figura\,anexada.}\\\sf pelo\,Teorema\,de\,Pit\acute agoras:\\\sf d^2=\ell^2+\ell^2\\\sf d^2=2\ell^2\\\sf d=\sqrt{2\ell^2}\\\sf d=\ell\sqrt{2}.\\\sf note\,que\,\ell\,\acute e\,o\,cateto\,oposto\\\sf ao\,\hat angulo\,de\,45^\circ\,d\,a\,hipotenusa.\\\sf a\,relac_{\!\!,}\tilde ao\,que\,envolve\,cateto\,oposto\,e\,hipotenusa\\\sf \acute e\,o\,seno.\\\sf portanto\\\sf sen(45^\circ)=\dfrac{\ell}{d}\end{array}}$

$\large\boxed{\begin{array}{l}\sf sen(45^\circ)=\dfrac{\backslash\!\!\!\ell}{\backslash\!\!\!\ell\sqrt{2}}=\dfrac{1}{\sqrt{2}}\\\sf racionalizando\,temos:\\\sf sen(45^\circ)=\dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf sen(45^\circ)=\dfrac{\sqrt{2}}{\sqrt{2^{\backslash\!\!\!2}}}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf sen(45^\circ)=\dfrac{\sqrt{2}}{2}}}}}\end{array}}$