In triangle $BCD$, $\angle C = 90^\circ$, $CD = 3$, and $BD = \sqrt{13}$. What is $\tan B$?
Question:
In triangle $BCD$, $\angle C = 90^\circ$, $CD = 3$, and $BD = \sqrt{13}$. What is $\tan B$?
Answer:
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\frac{3}{2}
In triangle $BCD$, $\angle C = 90^\circ$, $CD = 3$, and $BD = \sqrt{13}$. What is $\tan B$?
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