We can draw the following picture:
The sine function is given by
[tex]\begin{gathered} \sin \theta=\frac{Opposite}{hypotenuse} \\ \sin \theta=\frac{3}{\sqrt[]{13}} \end{gathered}[/tex]
Now, we know that
[tex]\begin{gathered} \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}\times1 \\ \text{and we can write 1 as } \\ 1=\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \text{then} \\ \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}\times\frac{\sqrt[]{13}}{\sqrt[]{13}} \end{gathered}[/tex]
but square root of 13 times square root of 13 is equal to 13, I mean
[tex]\sqrt[]{13}\times\sqrt[]{13}=13[/tex]
then, we have
[tex]\begin{gathered} \frac{3}{\sqrt[]{13}}=\frac{3}{\sqrt[]{13}}\times\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \frac{3}{\sqrt[]{13}}=\frac{3\sqrt[]{13}}{13} \end{gathered}[/tex]
then, an equivalent answer is
[tex]\sin \theta=\frac{3\sqrt[]{13}}{13}[/tex]
the cosine function is given by
[tex]\begin{gathered} \cos \theta=\frac{Adjacent}{\text{hypotenuse}} \\ \text{cos}\theta=\frac{2}{\sqrt[]{13}} \end{gathered}[/tex]
this answer can be rewritten as
[tex]\begin{gathered} \text{cos}\theta=\frac{2}{\sqrt[]{13}}\times1 \\ \text{cos}\theta=\frac{2}{\sqrt[]{13}}\times\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \text{cos}\theta=\frac{2\sqrt[]{13}}{13} \end{gathered}[/tex]
and the tangent funcion is given by
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \tan \theta=\frac{3}{2} \end{gathered}[/tex]
