solution: 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 $\tan A = \frac{\sin A}{\cos A} = \frac{2\cos A}{\cos A } =\boxed{2}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  In right triangle $ABC$ with $\angle B = 90^\circ$, we have $\sin A = 2\cos A$.  What is $\tan A$?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 2
extra_info.question: In right triangle $ABC$ with $\angle B = 90^\circ$, we have $\sin A = 2\cos A$.  What is $\tan A$?
extra_info.split: test