How Do I Solve This Vector Displacement Problem?

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Bensky
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Need help on a vector problem by tomorrow. :(

Homework Statement


A woman walks 277. m in the direction 41.6° east of north, then 171 m directly east. Find
(a) her resultant vector displacement from the starting point

There were other parts to the question which I got right, I can't seem to get this one.

Homework Equations



None except the pythagorean theorem:
In this case, D=[tex]\sqrt{Dx^2 + Dy^2}[/tex]


The Attempt at a Solution



D1X=cos(41.6°) * 277
D1X=207.1400711

D2X=171

D1Y=sin(41.6°) * 277
D1Y=183.9075609

D2Y=0

Dx=378.14
Dy=183.9075604

D=sqrt(378.14^2+183.9075609^2)
D=420.49 =~ 420.5

Ok, so I got 420.5 and plugged in it. Nope, the computer says it's wrong. I tried rounding it to 421 and that was wrong also. I don't think I made any math errors, what am I doing wrong?
 
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You used the wrong angle..problem says east of north but you used north of east...otherwise, your work looks good once you correct for the proper angle.
 
PhanthomJay said:
You used the wrong angle..problem says east of north but you used north of east...otherwise, your work looks good once you correct for the proper angle.

Now I'm confused, what is the difference between east of north and north of east? >_> Which angle would I use?

EDIT: The only other angle I can think of is 48.4 degrees, found by: 90+41.6+x=180
 
Last edited:
x degrees "east of north" means x degrees from north in a clockwise manner. x degrees "north of east" means x degrees from east in an anticlockwise manner. Unless x=45, the angles will not be the same.
 
Thanks for explaining that, I changed the angle and it worked. :)