In triangle $ABC,$ $\angle C = 90^\circ$ and $D$ is the midpoint of $\overline{AC}.$  If $\sin \angle DBC = \frac{3}{5},$ then find $\tan \angle ABC.$
Since $\sin \angle DBC = \frac{3}{5},$ we can assume that $CD = 3$ and $BD = 5.$  Then by Pythagoras, $BC = 4.$

[asy]
unitsize(0.5 cm);

pair A, B, C, D;

B = (0,0);
C = (4,0);
D = (4,3);
A = (4,6);

draw(A--B--C--cycle);
draw(B--D);

label("$A$", A, NE);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, E);

label("$3$", (C + D)/2, E);
label("$3$", (A + D)/2, E);
label("$4$", (B + C)/2, S);
label("$5$", interp(B,D,0.75), NW);
[/asy]

Since $D$ is the mid-point of $\overline{AC},$ $AD = 3.$  Hence,
\[\tan \angle ABC = \frac{6}{4} = \boxed{\frac{3}{2}}.\]