A watermelon seed has the following coordinates: x = -6.6 m, y = 2.2 m, and z = 0 m. Find its position vector as (a) a magnitude and (b) an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates (8.2 m, 0 m, 0 m), what is its displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis? A) Magnitude was found using (-6.6)^2+(2.2)^2=sqrt(48.4)=6.96m or 7.0m B) The angle was found using inverse tangent of (2.2/6.6)=18.4 degrees. I subtracted 18.4 from 180 to get 161.6. C) The magnitude was found using the displacement equation (Δx=X2-X1). 8.2m-(-6.6m)=14.8m Up until part D, all of the answers are correct in WileyPlus. I have tried multiple ways trying to find this angle however, no success. Perhaps I am looking past the basic point. Please help.