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    long3's Avatar
    long3 Posts: 1, Reputation: 1
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    #1

    May 8, 2008, 07:00 PM
    physics/trignometry 140N and 35 deg
    So it says that a vector 140N is oriented at 35 deg with the horizontal. What is the magnitude of the horizontal component?
    So I figured this is somewhat of a trig question with the vector being 140N and the theta or angle between this triangle being 35 degrees.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    May 9, 2008, 04:43 AM
    The horizontal component would be 140cos(35)

    The vertical 140sin(35)
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    iamthetman Posts: 105, Reputation: 8
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    #3

    May 9, 2008, 06:58 AM
    I agree with the horizontal component.

    The vertical component would be 140sin(35).
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    galactus Posts: 2,271, Reputation: 282
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    #4

    May 9, 2008, 10:30 AM
    DUH, that's what I meant. Mistyped. Thanks for the catch
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    iamthetman Posts: 105, Reputation: 8
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    #5

    May 9, 2008, 11:03 AM
    I thought so.
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