Respuesta :
[tex]\bf (\stackrel{a}{4}~,~\stackrel{b}{-\sqrt{33}})\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{4^2+(-\sqrt{33})^2}\implies c=\sqrt{16+33}\implies c=\sqrt{49}
\\\\\\
cos(\theta )=\cfrac{\stackrel{adjacent}{4}}{\stackrel{hypotenuse}{\sqrt{49}}}[/tex]
and now, let's rationalize the denominator,
[tex]\bf \cfrac{4}{\sqrt{49}}\cdot \cfrac{\sqrt{49}}{\sqrt{49}}\implies \cfrac{4\sqrt{49}}{(\sqrt{49})^2}\implies \cfrac{4\sqrt{49}}{49}[/tex]
and now, let's rationalize the denominator,
[tex]\bf \cfrac{4}{\sqrt{49}}\cdot \cfrac{\sqrt{49}}{\sqrt{49}}\implies \cfrac{4\sqrt{49}}{(\sqrt{49})^2}\implies \cfrac{4\sqrt{49}}{49}[/tex]
from the coordinates this angle is in the 4th quadrant
Hypotenuse = sqrt (4^2 + 33) = sqrt 49 = 7
cos θ = 4/7
Hypotenuse = sqrt (4^2 + 33) = sqrt 49 = 7
cos θ = 4/7
