[tex]\bf tan(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{adjacent}{4}}\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{3^2+4^2}\implies c=5
\\\\\\
sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{hypotenuse}{5}}[/tex]
bear in mind that, though the square has two valid roots, one negative and one positive, the hypotenuse is just a radius distance, and therefore is never negative.
