Vector A is the length 2cm and 60° above the x-axis in the first quadrant. Vector B is of length 2cm and 60° below the x-axis in the fourth quadrant. The sum vector A+B is a vector of magnitude-

2 along +X axis

|A+B|^2 = 2^2 + 2^2 - 2(2)(2)cos 120°

To find the magnitude of the sum vector A+B, we first need to find the components of vector A and vector B.

For vector A:
Since it is 60° above the x-axis in the first quadrant, we can represent it as A = (2cos(60°), 2sin(60°)). Using the trigonometric functions, we find that cos(60°) = 0.5 and sin(60°) = √3/2.
So, the components of vector A are A = (2 * 0.5, 2 * √3/2) = (1, √3).

Similarly, for vector B:
Since it is 60° below the x-axis in the fourth quadrant, the angle it makes with the positive x-axis is 180° - 60° = 120°. Thus, we can represent it as B = (2cos(120°), 2sin(120°)). Using the trigonometric functions, we find that cos(120°) = -0.5 and sin(120°) = √3/2.
So, the components of vector B are B = (2 * -0.5, 2 * √3/2) = (-1, √3).

Now, let's add the components of vectors A and B:
A + B = (1, √3) + (-1, √3) = (0, 2√3).

To find the magnitude of the sum vector, we use the formula:
Magnitude = √(x² + y²).

Applying this formula to the sum vector, we have:
Magnitude = √((0)² + (2√3)²) = √(0 + 12) = √12.

Therefore, the magnitude of the sum vector A+B is √12 units.