# Solving vectors using component method

1. Jul 5, 2011

### yety124

1. The problem statement, all variables and given/known data
Add the following vectors using component method:
Vector A=175km; 30 degrees north of east
Vector B=153km; 20 degrees west of north
Vector C=195km; West
a)Find x and y component of each vector
b)Solve for the magnitude of resultant vector
c) Solve for direction(angle)

2. The attempt at a solution
So i these are what i got
Vector A: x=175cos 30....x=151.56 y=175sin 30......y=87.5
Vector B: x=153cos 20....x=-143.77 y=153sin 20.....y=52.33
(are these correct?)
so after getting this i added all the x's and y's but then i realized that the question had said that vector C goes 195km west, so i now got confused what to do next.

2. Jul 5, 2011

### Staff: Mentor

There is nothing special about "in a westerly direction". A westerly vector contributes a component D.cos 180o in the x-direction, and D.sin 0o in the y-direction.

3. Jul 5, 2011

### yety124

well that was stupid of me, thanks for the reply finally solved it :D

4. Jul 5, 2011

### HallsofIvy

Staff Emeritus
If $\theta$ is measured counterclockwise from the positive x-axis then a vector of length r and angle $\theta$ has components $rcos(\theta)$ and $r sin(\theta)$. Strictly speaking, you are free to choose the "positive x-axis" any way you want as long as you are consistent but the usual convention is that the positive x-axis points East.

For the first vector you are given that r= 175km and the directon is 30 degrees north of east. "north of east" is counterclockwise from east so the angle is $\theta= 30$ degrees.

For the second vector you are given that r= 153km and the direction is 20 degrees west of north. West is clockwise of north but north itself is 90 degrees clockwise of east. The angle is $\theta= 90+20= 110$ degrees.

For the third vector you are given that r= 195km and the direction is west. West is exactly opposite east so the angle is $\theta= 180$ degrees.