# Angle of Vector

1. Sep 14, 2010

### delfam

1. The problem statement, all variables and given/known data
car travels 18.2km due north then 43.2km at 53 degrees west of north. Find magnitude of car's resultant displacement.

b.) calculate the direction of the car's resultant displacement, measured counter-clockwise from the northerly direction.

2. Relevant equations

3. The attempt at a solution
Ax = 18.2sin(90) =18.2 Ay = 18.2cos(90) = 0
Bx = 43.2sin(143) = 26 By = 43.2cos(143) = -34.501
Cx = 44.2 Cy = -34.501

44.2^2 + -34.501^2 = 3143.82, square root of that = 56.07

arctan(-34.504/44.2) = -37.98 degrees

so part b says counter-clockwise from north, so wouldn't it be from 90 degrees so should the angle be 127.98? I entered that as an asnwer and it's wrong so not sure what I did wrong.

Last edited: Sep 14, 2010
2. Sep 14, 2010

### MHrtz

when you did Bx and By, where did you get 43.2 and 143 degrees?

3. Sep 14, 2010

### delfam

typo, it was due north 43.2km not 432, the 143 degrees is when it says 53 degrees west of north, north is 90, so 90 + 53 = 143. I already got that part right, I just don't understand the wording on part b.

4. Sep 14, 2010

### MHrtz

recheck your math for Bx and By. Bx should be 260 and By should be -345. Also, you're using the Pythagorean theorem when you should be using law of cosines. the angle across from the resultant vector should be 90 + (90 - 53) = 127. In other words, it's 53 degrees from the vertical not the horizontal.