Recent content by Koko23

Ax=-6.3 sin 23 = -2.46; Ay=6.3 cos 23 = 5.8; Bx=5.7 cos 34 =4.73; By=5.7 sin 34 = 3.19 Sqrt((Bx-Ax)^2 + (By-Ay)^2) = 7.7 units for the magnitude of C. For the angle measured from the x-axis to vector c, I got -20 which is the angle 20 degrees below the x-axis to vector c using tan.

This is the diagram for the problem. I drew the arrow pointing downward since it was vector B - vector A.

Ax=6.3 cos 23; Ay=-6.3 sin 23; Bx= 5.7 cos 34; By=5.7 sin 34. Is this correct to calculate vector C magnitude which I got 7.7 units. Also is vector C in quadrant IV? I am not sure how to calculate the angle part of this question.