Vector Addition Problem: How Far and In What Direction Did the Hiker Travel?

[Elysian]

Homework Statement

[PLAIN]http://img6.imageshack.us/img6/4057/pictureforforums.png
A hiker walks 8.33 km 30 degrees N of E then 4.97 km 45 degrees N of W and finally 7.32 km 60 degrees S of E. How far and in what direction is the hiker from the original position (A) to (B)

Homework Equations

Not sure if any, I used Law of Sines and Cosines

The Attempt at a Solution

As you can see in the picture, I subtracted 4.97 km from 7.32 km to get 2.35 km, I did this because N of W and S of E are opposites, not sure of right on that part though. So then I get a triangle with lenghts 8.33, and 2.35, and an angle of 30, Using Law of Sines

\stackrel{8.33}{sin(30)} = \stackrel{2.35}{sin(b)}

which gives B = 8.10898 degrees,

angles A B and C have to add up to 180 so C is 141.891

Law of Cosines is

c = \sqrt{8.33^2 + 2.35^2 - 2(8.33*2.35)cos141.891} = 10.2819 km

 

[PCSL]

Solve by breaking up into components - law of cosines is not needed.