In triangle $ABC,$ $\angle C = 90^\circ$ and $D$ is the midpoint of $\overline{AC}.$  If $\sin \angle DBC = \frac{3}{5},$ then find $\tan \angle ABC.$
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You are a helpful assistant.<|im_end|>
<|im_start|>user
In triangle $ABC,$ $\angle C = 90^\circ$ and $D$ is the midpoint of $\overline{AC}.$  If $\sin \angle DBC = \frac{3}{5},$ then find $\tan \angle ABC.$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
\frac{3}{2}
In triangle $ABC,$ $\angle C = 90^\circ$ and $D$ is the midpoint of $\overline{AC}.$  If $\sin \angle DBC = \frac{3}{5},$ then find $\tan \angle ABC.$
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