\frac{\sqrt{65}}{9}
Question:
In right triangle $BCD$ with $\angle D = 90^\circ$, we have $BC = 9$ and $BD = 4$.  Find $\sin B$.
Answer:
Let's think step by step.

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\frac{\sqrt{65}}{9}
In right triangle $BCD$ with $\angle D = 90^\circ$, we have $BC = 9$ and $BD = 4$.  Find $\sin B$.
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