# Homework Help: Vector treasure map problem

1. Sep 13, 2007

### eeriana

1. The problem statement, all variables and given/known data
While following a treasure map, you start at an old oak tree. You first walk 825 m directly south, then turn and walk 1.25 km at 30degrees west of north, then 1.00 km 40.0 degrees north of east where you find a treasure. To return to the oak tree, in what direction would you walk and how far?

2. Relevant equations

3. The attempt at a solution
I thought that I did everything right, but can't seem to get the answer that they have in the back of the book. Any help would be appreciated.

R=A+B+C.
A = 270 degrees B= 120 degrees C = 50 degrees

Ax = Acos$$\Theta$$= (825m)(cos 270) = 0
Ay = Asin $$\Theta$$= (825)(sin270) =-825m
Bx= (1250m)(cos 120) = -625m
By= (1250) (sin 120) = 1082.53m
Cx=(1000)(cos50) = 642.78m
Cy=(1000)(cos 50) = 766m

Rx= 17.78m Ry = 1023.53 R= $$\sqrt{}(17.78^2)+(1023.53^2)$$ = 1023.68m

arctan (1023.53/17.78) = 89 degrees

The answer I come up with is 1023.7 m at 1 degree west of south.

The answer in the book is 911 m at 8.9 degrees w of s.

Thanks for the help!!!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 13, 2007

### learningphysics

Your third angle should be 40 not 50... is use cos40 not cos50 etc...

3. Sep 14, 2007

### eeriana

Forgive me, but could you explain why. Perhaps I am not understanding the angle calculation, but I thought it was counterclockwise from the + x axis towards the + y axis and I thought that with it being 40 degrees north of east, it would be 90-40. I am sorry if this sounds like a stupid question, but I am trying to understand where I am going wrong so that I can understand what to do the next time...

Thank you :)

Amy

4. Sep 14, 2007

### learningphysics

If you go counterclockwise from the +x axis towards the y-axis... that's the same as going north of east... The angle east of north is 50... the angle north of east is 40.

going north from east... is the same as going counterclockwise towards the y-axis... because you are going up from the positive x-axis...

Last edited: Sep 14, 2007
5. Sep 14, 2007

### eeriana

I think I see what you're saying.