Question

16. (UF-AL) A +1sen300°/tg540+ cos(-120°) é igual a:
( ) - √3/3
( )√3/4
( )2-√3/4
( )2+√3
( ) -2+√3

Answer 1

$\Large\boxed{\begin{array}{l}\sf sen(300^\circ)=-sen(60^\circ)=-\dfrac{\sqrt{3}}{2}\\\sf tg(540^\circ)=tg(180^\circ)=0\\\sf cos(-120^\circ)=cos(240^\circ)=-cos(60^\circ)=-\dfrac{1}{2}\\\sf\dfrac{1+sen(300^\circ)}{tg(540^\circ)+cos(-120^\circ)}=\dfrac{1-\frac{\sqrt{3}}{2}}{0-\frac{1}{2}}\\\\\sf\dfrac{\frac{2-\sqrt{3}}{\diagdown\!\!\!2}}{-\frac{1}{\diagdown\!\!\!2}}=-2+\sqrt{3}\end{array}}$