Vec A = (1.28,2.26) Vec B (4.24,-3.72) Find CCW Angle that a+b makes on xaxis?

Question

Looking for guidance.. I've arrived at the magnitude is 5.71cm, however I can't seem to visual the interval of the angle.

The answer has to be in the form of some interval ... e.g. [0,360deg] ...??

0,-14.8 and -14.8,0 retrieve the results of "off by an additive constant"...what is it that I didn't factor in.. components, magnitude..atan* ??

The answer is 345 deg! -- How?

Answer 1

A + B = (5.52, -1.46)
Magnitude = 5.7098
Angle = -0.25857 rad = -14.815 deg
This angle, being negative, represents clockwise rotation from the x axis. The A + B vector is in the 4th (lower right) quadrant. It shouldn't be hard to visualize. From (0,0), you go right 5.52 on the x axis, then down -1.46 on the y axis. The line drawn from (0,0) to this point makes a small angle with the x axis. (Less than 45 deg since 1.46 < 5.52)
EDIT: OK, you need it in interval form, the angle covers the interval 0 to -14.815 deg.
EDIT: It's 345 deg because you can add 360 deg to an angle and it's still the same angle plus exactly one rotation. Also if the answer is exactly 345 deg, they're rounding it; 360-14.815=345.185 deg is more correct..

Answer 2

since you were able to get the magnitude of the resultant vector.. the angle is just given by...

angle = arctan (Ry / Rx)

where Ry is the y component of the resultant vector and Rx is the x component.