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

<think>
<Plan_1>
Okay, let's try to figure out this problem. So, we have a right triangle ABC with a right angle at B. That means angle A and angle C must add up to 90 degrees since the total sum of angles in a triangle is 180. The problem states that sin A equals 2 cos A, and we need to find tan A. Hmm, let's start by recalling some trigonometric identities and relationships in right triangles.

First, in a right triangle, the sine of an angle is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. Since angle B is 90 degrees, angles A and C are the other two angles, each less than 90.

<prompt_1.1> Set up the relationships for sin A, cos A, and tan A in terms of the sides of the triangle. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's focus on angle A. Let me denote the sides: opposite to angle A is BC, adjacent is AB, and the hypotenuse is AC. So, sin A = BC/AC and cos A = AB/AC. The equation given is sin A = 2 cos A, which translates to BC/AC = 2*(AB/AC). Since the denominators are the same, we can equate the numerators: BC = 2*AB. So, the length of BC is twice the length of AB.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But wait, maybe I should approach this using trigonometric identities instead of the sides. The equation given is sin A = 2 cos A. If I divide both sides by cos A, assuming cos A isn't zero, which it can't be in a right triangle because angle A is acute. So, dividing both sides by cos A gives tan A = 2. Wait, that seems straightforward. So, tan A is 2?

Wait, let me check that again. Starting from sin A = 2 cos A. If I divide both sides by cos A, I get sin A / cos A = 2, which is tan A = 2.

But maybe the problem is designed to test if I go through the sine and cosine definitions using the triangle sides. Let me confirm using both methods to ensure consistency.

First, using the identity approach: sin A = 2 cos A. Dividing both sides by cos A gives tan A = 2. So, the answer is 2.

But let's verify with the triangle sides. Let's assign variables.

<prompt_2.1> Verify the result using the triangle sides by assigning variables and solving for the sides. </prompt_2.1>

Alternatively, maybe using the identity sin^2 A + cos^2 A = 1. Since sin A = 2 cos A, substitute into the identity: (2 cos A)^2 + cos^2 A = 1 => 4 cos^2 A + cos^2 A = 1 => 5 cos^2 A = 1 => cos^2 A = 1/5 => cos A = 1/√5 (since cosine is positive in acute angles), so sin A = 2/√5. Then tan A = sin A / cos A = (2/√5)/(1/√5) = 2.
</Plan_2>

<Execution_2>
<execution_2.1>
Let’s let AB be the adjacent side, BC be the opposite side, and AC be the hypotenuse. Then, as per the equation sin A = 2 cos A, which is BC/AC = 2*(AB/AC). The AC cancels out, so BC = 2 AB. So, the ratio of opposite over adjacent is BC/AB = 2, which is tan A. Therefore, tan A is 2.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Therefore, multiple approaches confirm that tan A is 2.

Wait, but let me just visualize the triangle to ensure there's no confusion with the labels.

<prompt_3.1> Visualize the triangle to ensure there is no confusion with the labels and confirm the result. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
In triangle ABC, right-angled at B. So, points A, B, C with B being the right angle. So, sides AB and BC are the legs, and AC is the hypotenuse. Angle A is at vertex A, so the sides opposite and adjacent to angle A would be BC and AB, respectively. So, yes, tan A is BC/AB, which we found to be 2.
</execution_3.1>
</Execution_3>

<Final_answer>
Therefore, the value of $\tan A$ is $\boxed{2}$.
</Final_answer>
</think>