SInce,
[tex]\begin{gathered} \sin B=\frac{opposite}{hypotenuse} \\ \text{opposite}=3 \\ \text{hypotenuse}=8 \end{gathered}[/tex]
Therefore using Pythagoras theorem, we have,
[tex]x=\sqrt[]{8^2-3^2}=\sqrt[]{64-9}=\sqrt[]{55}[/tex]
Therefore, tan B can be calculated as,
[tex]\tan \text{ B=}\frac{opposie}{adjacent}=\frac{3}{\sqrt[]{55}}[/tex]
