[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
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cos(\theta )=\cfrac{2\sqrt{10}}{7}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\impliedby \textit{now, let's find the opposite side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm \sqrt{c^2-a^2}=b\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}[/tex]
[tex]\bf \pm \sqrt{7^2-(2\sqrt{10})^2}=b\implies \pm\sqrt{49-(2^2\sqrt{10^2})}=b
\\\\\\
\pm\sqrt{49-(4\cdot 10)}=b\implies \pm\sqrt{9}=b\implies \pm 3=b
\\\\\\
\textit{no quadrant is given for the angle, so, we can just assume is the +3}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad \qquad sin(\theta )=\cfrac{3}{7}[/tex]
