trig

B

                sqrt 14              

C       3             D

BC =  sqrt [ (sqrt 14)^2 - 3^2 ]   = sqrt  5

tan B  =   CD  /   BC  =     3 /  sqrt 5  =   3sqrt (5)  /  3

By the Pythagorean Theorem, $BC= \sqrt 5$. 

Because $\tan = { \text{opposite} \over \text{adjacent}}$, $\tan B = {3 \over \sqrt 5 }$, which can be rationalized to $\color{brown}\boxed{3 \sqrt 5 \over 5}$