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Can Vectors with an angle 180(degrees+) have a negative magnitude?

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Can vectors with 180+(degrees) have a negative magnitude? I'm trying to find components of a vector that is going 260 km, 48 (degrees) south of east. So I'm confused whether the 260 is positive or not because of the -48 degrees.


    2. Relevant equations
    ([itex]\vec{V}[/itex]) (sin[itex]\theta[/itex])
    ([itex]\vec{V}[/itex]) (cos[itex]\theta[/itex])

    3. The attempt at a solution
    (-260km)(sin(48)) or (260km)(sin(48)) =[itex]\vec{V}[/itex][itex]_{y}[/itex]

    (-260km)(cos(48)) or (260km)(cos(48)) =[itex]\vec{V}[/itex][itex]_{x}[/itex]

    Which one??
     
  2. jcsd
  3. Sep 16, 2011 #2
    sin(x) = -sin(-x)

    cos(x) = cos(-x)

    Hope that helps.

    Just use the convention that counterclockwise is postive and clockwise is negative in terms of measuring an angle.
     
    Last edited: Sep 16, 2011
  4. Sep 16, 2011 #3
    Thank you this helped so much! But um just a quick question, why is sin negative? o_O
     
  5. Sep 16, 2011 #4
    Oh wait is this from cos, sin
    and since it is in the 4th quadrant, y is negative (sin) and x is positive (cos)
     
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