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

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
($\vec{V}$) (sin$\theta$)
($\vec{V}$) (cos$\theta$)

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

(-260km)(cos(48)) or (260km)(cos(48)) =$\vec{V}$$_{x}$

Which one??
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 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.

 Quote by WatermelonPig 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.
Thank you this helped so much! But um just a quick question, why is sin negative? O.o

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

 Quote by SaltyBriefs Thank you this helped so much! But um just a quick question, why is sin negative? O.o
Oh wait is this from cos, sin
and since it is in the 4th quadrant, y is negative (sin) and x is positive (cos)

 Tags components, homework, signs, trigonometry, vectors

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