[tex]\bf (\stackrel{a}{3}~,~\stackrel{b}{-2})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c = \sqrt{3^2+(-2)^2}\implies c=\sqrt{9+4}\implies c=\sqrt{13} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf cos(\theta )\implies \cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{\sqrt{13}}}\implies \cfrac{3}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{3\sqrt{13}}{13} \\\\\\ sin(\theta )\implies \cfrac{\stackrel{opposite}{-2}}{\stackrel{hypotenuse}{\sqrt{13}}}\implies \cfrac{-2}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{-2\sqrt{13}}{13} \\\\\\ cot(\theta )\implies \cfrac{\stackrel{adjacent}{3}}{\stackrel{opposite}{-2}}[/tex]
