1. The problem statement, all variables and given/known data A car travels 20 km due north and then 35 km in a direction 60 degree west of north. Find the magnitude and direction of the car's resultant displacement. 2. Relevant equations R = sqrt(A^2 + B^2) Ax = cosA, Ay = sinA tan = y/x 3. The attempt at a solution This question was introduced and solved in the textbook before the author introduced Components of a Vector and Unit Vectors (i, j, z; Ax, Ay, Az). I came back to this question as an attempt. The author used law of cosines in his deomonstration, as well as sin beta to solve the angle (direction). He gave R = 48.2km at 38.9 I tried this problem with component method. I drew a picture and I started the problem by listing components: let the north 20km = A, 35km = B and their resultant = R Ax = 0 (by all means, cos90 * 20 gives zero anyway) Ay = sin90 * 20 = 20 Bx = cos60 * 35 = 17.5 By = sin60 * 35 = 30.3 ** --> = vector --> R = (Ax + Bx)i + (Ay + By)j --> R x = Ax + Bx (17.5) --> Ry = Ay + By (50.3) and --> R = sqrt (Rx^2 + Ry^2) In the end, R I got ~ 53 km, and for the degree, I got tan (Ry/Rx) = 70.7, but since the picture shows the direction is beyond 90, I say 180 - 70.7 = 109.3 Where did I do wrong? Thanks.