Answer :
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{hypotenuse}{4}}\impliedby \textit{now let's find the \underline{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}[/tex]
[tex]\bf \pm\sqrt{4^2-3^2}=a\implies \pm\sqrt{16-9}=a \\\\\\ \pm\sqrt{7}=a\implies \stackrel{\textit{in the I Quadrant}}{+\sqrt{7}=a}\qquad therefore \\\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{3}{\sqrt{7}} \\\\\\ \textit{and rationalizing the denominator}\qquad tan(\theta)=\cfrac{3\sqrt{7}}{7}[/tex]
[tex]\bf \pm\sqrt{4^2-3^2}=a\implies \pm\sqrt{16-9}=a \\\\\\ \pm\sqrt{7}=a\implies \stackrel{\textit{in the I Quadrant}}{+\sqrt{7}=a}\qquad therefore \\\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{3}{\sqrt{7}} \\\\\\ \textit{and rationalizing the denominator}\qquad tan(\theta)=\cfrac{3\sqrt{7}}{7}[/tex]