Respuesta :

[tex]\bf \begin{array}{clclll} (\sqrt{33}&,&-4)\\ x&&y \end{array}\qquad \qquad tan(\theta)=\cfrac{y}{x}\implies tan(\theta )=\cfrac{-4}{\sqrt{3}} \\\\\\ \textit{now, let's rationalize the denominator} \\\\\\ \cfrac{-4}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{-4\sqrt{3}}{(\sqrt{3})^2}\implies \boxed{\cfrac{-4\sqrt{3}}{3}}[/tex]