The triangle is shown below:

[asy]
pair A,B,C;
A = (0,0);
B = (5,0);
C = (5,10);
draw(A--B--C--A);
draw(rightanglemark(C,B,A,16));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
[/asy]

We have $\sin A = \frac{BC}{AC}$ and $\cos A = \frac{AB}{AC}$, so $\sin A = 2\cos A$ gives us $\frac{BC}{AC} = 2\cdot\frac{AB}{AC}$.  Multiplying both sides by $AC$ gives $BC = 2AB$, so $\frac{BC}{AB} = 2$.  Finally, we have $\tan A = \frac{BC}{AB} = \boxed{2}$.

We also could have noted that $\tan A = \frac{\sin A}{\cos A} = \frac{2\cos A}{\cos A } =\boxed{2}$.
Question:
In right triangle $ABC$ with $\angle B = 90^\circ$, we have $\sin A = 2\cos A$.  What is $\tan A$?
Answer:
Let's think step by step.

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2
In right triangle $ABC$ with $\angle B = 90^\circ$, we have $\sin A = 2\cos A$.  What is $\tan A$?
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