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Finding angles of vectors

  1. Oct 9, 2007 #1


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    1. The problem statement, all variables and given/known data
    What is the y-component of a vector in the x-y plane whose magnitude is 86.4 and whose x-component is 62.3?

    2. Relevant equations
    a^2 + b^2 = c^2
    Sin-1= Vy/|v|

    3. The attempt at a solution
    a^2 + b^2 = c^2
    And 59.9 was right for that, but

    i used Sin-1= Vy/|v| for the angle and got:
    Sin-1= 59.86/ 86.4
    = 43.85°
    but that wasn't right.
    Can anyone offer me some insight?
  2. jcsd
  3. Oct 9, 2007 #2
    Hmm strange. Perhaps the vector is not in the first quadrant if the angle 43.85 degrees is not right.

    See what is the correct answer if it is supplied and work your way back to understand it.
    From the question it looks like X is positive so the angle is either in the first or fourth quadrant.
  4. Oct 9, 2007 #3


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    Oh, it also said to find the angle with respect to the x -axis, if that clarifies. So if it is in the fourth quadrant the angle would be negative?
    Last edited: Oct 9, 2007
  5. Oct 9, 2007 #4
    Yes if you are looking clockwise and if the angle is indeed in the fourth quadrant which is the most likely answer your calculator will give you.

    If you still not understand, post the answer your book tells you and I will try to decode it for you.
  6. Oct 9, 2007 #5


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    Well, we submit our answers online and it only tells us if it's right or wrong. But -43.85 was also incorrect, so should I be adding like 90° or 180° to it? As if it was trying to find the angle around it, and not the angle itself?
  7. Jul 23, 2008 #6
    Two vectors have a magnitude of 86.4 and an x coomponent of 62.3:

    In the first quadrant (62,3, 59.9)
    In the fourth quadrant (62.3, -59.9)

    Their polar coordinates are:
    In the first quadrant (86.4, 43.86 degrees)
    In the fourth quadrant (86.4, -43.86 degrees)
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