Vector A has a magnitude of 14.0m and is angled 38.0 degrees counterclockwise from the +x direction. Vector C has a magnitude of 13.2m and is angled at 22.4 degrees counterclockwise to the -x direction. If vector A + vector B = vector C what is (a) the magnitude (b) and angle - relative to +x direction - of vector B? State your angle as a positive number. Answers: (a) 26.9m (b) 210 degrees Relevant equations: I will use vA as a shorthand to represent vector A and ||vA|| to represent the magnitude of vector A. Ax = ||vA||cos(theta) Ay = ||vA||sin(theta) ||vA|| = sqrt(Ax^2 + Ay^2) --------------------------------- My reasoning was that if I could find the vectors of vA and vC I could subtract vA from vC to obtain the vector vB. vA + vB = vC vB = vC - vA = <Cx-Ax, Cy-Ay> = <Bx, By> From there I would use the equation for the magnitude of the vector to obtain the answer for part a. ||vB|| = sqrt(Bx^2 + By^2) However, when I plug in the numbers I keep getting the incorrect answer, 23.5m. Ax = 11.0m Ay = 8.62m vA = <11.0m, 8.62m> Cx = -12.2m Cy = 5.03m vC = <-12.2m, 5.03m> vB = <-12.2m-11.0m, 5.03m-8.62m> = <-23.2m, -3.59m> ||vB|| = sqrt((-23.2^2 + (-3.59)^2)) = sqrt(551.13) = 23.5m I haven't attempted part (b) yet due to the fact that there is an error to the first part of the question. What did I do wrong? I felt fairly confident in what I was doing and have been stumped on this question for nearly an hour. Thank you in advance for reading my problem.