so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant
[tex]\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{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{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf cos(\theta )=\cfrac{\stackrel{adjacent}{9}}{\stackrel{hypotenuse}{\sqrt{90}}}\implies \stackrel{\textit{rationalizing the denominator}}{\cfrac{9}{\sqrt{90}}\cdot \cfrac{\sqrt{90}}{\sqrt{90}}\implies \cfrac{9\sqrt{90}}{90}}\implies \cfrac{\sqrt{90}}{10}\implies \cfrac{3\sqrt{10}}{10}[/tex]
