In triangle $ABC,$ $\angle C = 90^\circ$ and $D$ is the midpoint of $\overline{AC}.$  If $\sin \angle DBC = \frac{3}{5},$ then find $\tan \angle ABC.$
Question:
In triangle $ABC,$ $\angle C = 90^\circ$ and $D$ is the midpoint of $\overline{AC}.$  If $\sin \angle DBC = \frac{3}{5},$ then find $\tan \angle ABC.$
Answer:
Let's think step by step.

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\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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