[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{2}{7}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{7^2-2^2}=a\implies \pm\sqrt{45}=a\implies \pm 3\sqrt{5}=a[/tex]
but.... which is it? the + or the -? well, we know that tan(θ) > 0, is another way to say that the tangent of the angle is positive, now, for the tangent to be positive, since it's opposite/adjacent both opposite and adjacent have to be the same exact sign, now, we know the opposite is +2, so that means the adjacent has to be the same sign, thus is the positive version 3√(5)
thus [tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad \qquad cos(\theta )=\cfrac{3\sqrt{5}}{7}[/tex]
