<Question>
In a convex quadrilateral, the measure of the largest angle is twice the measure of the smallest angle, and the other two angles are both right angles. How many degrees are in the largest angle?
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the measure of the largest angle in a convex quadrilateral where the largest angle is twice the smallest one, and the other two angles are right angles (90 degrees each). Hmm, let's start by recalling some basics about quadrilaterals.

<prompt_1.1> Recall the basic properties of quadrilaterals and set up an equation for the sum of the angles. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, I remember that the sum of the interior angles of any convex quadrilateral is 360 degrees. That's a key point. So, if I can figure out the measures of all four angles, they should add up to 360. The problem states that two of the angles are right angles, which are 90 degrees each. Let's denote the smallest angle as, say, x degrees. Then the largest angle, which is twice the smallest, would be 2x degrees. The other two angles are right angles, so 90 degrees each. So, adding them up: x + 2x + 90 + 90 = 360.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me check that equation again.

<prompt_2.1> Verify the equation set up in the above plan. </prompt_2.1>
-----

But wait, just to be thorough, are there any other constraints or properties of a convex quadrilateral that I need to consider here? For example, in a convex quadrilateral, all interior angles must be less than 180 degrees.

<prompt_2.2> Ensure that all angles in the quadrilateral are valid for a convex shape. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
That seems right. Combining like terms, that would be 3x + 180 = 360. Then, subtracting 180 from both sides gives 3x = 180. Dividing both sides by 3, x = 60. Therefore, the smallest angle is 60 degrees, and the largest angle is 2x = 120 degrees.
</execution_2.1>

<execution_2.2>
Since the largest angle is 2x, and we found x=80, so 2x=160, which is less than 180, so that's okay. The right angles are 90, which are also less than 180. The smallest angle is 80, which is also fine. So all angles are less than 180, so the quadrilateral is indeed convex.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, hold on. Let me make sure I didn't skip any steps or make any mistakes here. Let's go through it again.

<prompt_3.1> Re-evaluate the steps to verify the solution. </prompt_3.1>
-----

But let me just think again: the problem says "the measure of the largest angle is twice the measure of the smallest angle," and the other two are right angles. So, if the two right angles are 90 each, then the other two angles must be x and 2x. Since in a convex quadrilateral, all angles must be less than 180, so 2x < 180, so x < 90.
But here, x is 80, which is less than 90, so that's okay. So, all angles would be 80, 90, 90, 160. Adding up to 80+90+90+160=420?

<prompt_3.2> Re-calculate and verify the sum of the angles to ensure correctness. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
The quadrilateral has four angles: two right angles (90° each), one smallest angle (x), and one largest angle (2x). Summing them: x + 2x + 90 + 90. That's 3x + 180. Total sum must be 360, so 3x = 180. So x = 60. Then 2x = 120.
</execution_3.1>

<execution_3.2>
Wait, no, 80+90 is 170, plus another 90 is 260, plus 160 is 420. Wait, that can't be. Wait, 80+90+90+160= 80+90=170, 170+90=260, 260+160=420. That's 420, which is way more than 360. Wait, that's a problem. Did I make a mistake here?
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
Wait, hold on. Wait, no, if x is 80, then 2x is 160. Then adding 80 + 90 + 90 + 160: 80+160 is 240, 90+90 is 180, 240+180 is 420. Wait, that's 420. That's not 360. So that's a problem.

<prompt_4.1> Identify and correct the error in the calculation. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Wait, but earlier when I solved the equation, I had x + 2x + 90 + 90 = 3x + 180 = 360, leading to x = 60, 2x = 120. Then the angles would be 60, 120, 90, 90. Adding those: 60+120=180, 90+90=180, total 360. That's correct.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But in my previous calculation, I mistakenly used x as 80, which was a miscalculation. Wait, so where did that 80 come from? Wait, no. Let me check again.

<prompt_5.1> Correct the error and verify the solution again. </prompt_5.1>
-----

But just to be absolutely thorough, let's consider if there's another interpretation of the problem. For example, maybe "the largest angle is twice the smallest angle" could mean that there are multiple angles, but the largest is twice the smallest, and the other two are right angles. But in a quadrilateral, there are only four angles. So if two are right angles (90), and the other two are x and 2x, that's the only possibility. So, yes, x=60, 2x=120.
</Plan_5>

<Execution_5>
<execution_5.1>
If the two right angles are 90 each, then the other two angles are x (smallest) and 2x (largest). So total sum is x + 2x + 90 + 90 = 3x + 180. Set equal to 360: 3x = 180, so x = 60. Then 2x = 120. So the angles are 60, 120, 90, 90. Adding them up: 60+120=180, 90+90=180, total 360. That works. So the largest angle is 120 degrees.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the measure of the largest angle in the convex quadrilateral is $\boxed{120^\circ}$.
</Final_answer>
</think>