In triangle $ABC,$ $\angle C = 90^\circ$ and $M$ is the midpoint of $\overline{BC}.$  If $\sin \angle BAM = \frac{1}{3},$ then what is $\sin \angle BAC$?
Question:
In triangle $ABC,$ $\angle C = 90^\circ$ and $M$ is the midpoint of $\overline{BC}.$  If $\sin \angle BAM = \frac{1}{3},$ then what is $\sin \angle BAC$?
Answer:
Let's think step by step.

user
\frac{\sqrt{6}}{3}
In triangle $ABC,$ $\angle C = 90^\circ$ and $M$ is the midpoint of $\overline{BC}.$  If $\sin \angle BAM = \frac{1}{3},$ then what is $\sin \angle BAC$?
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