so the point is at (2,3) namely x = 2, and y = 3, or one could say a = 2 and b = 3,
[tex]\bf (\stackrel{a}{2}~~,~~\stackrel{b}{3})\impliedby \textit{first off, 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{2^2+3^2}\implies c=\sqrt{4+9}\implies c=\sqrt{13}[/tex]
bearing in mind that, the hypotenuse is just the radius unit in the angle, and therefore is never negative.
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\quad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------[/tex]
[tex]\bf sin(\theta)=\cfrac{3}{\sqrt{13}}\impliedby \textit{let's rationalize the denominator}
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\cfrac{3}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{3\sqrt{13}}{\sqrt{13^2}} \implies \cfrac{3\sqrt{13}}{13}
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cos(\theta)=\cfrac{2}{\sqrt{13}}\impliedby \textit{let's rationalize the denominator again}
\\\\\\
\cfrac{2}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{2\sqrt{13}}{\sqrt{13^2}}\implies \cfrac{2\sqrt{13}}{13}
\\\\\\
tan(\theta)=\cfrac{3}{2}[/tex]
