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Math Help - Trig Help

  1. #1
    Newbie
    Joined
    Sep 2005
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    2

    Trig Help

    Hi, I need help finding the sin of the angle alpha in standard postion whose terminal side contains the given point. (-4,-6).

    Here's what I have so far.

    r = sqrt{(-4)^2 + (-6)^2} = sqrt{52}
    x = -4, y= -6, r= sqrt{52}

    sin(alpha) = (-6/sqrt(52))
    = (-6/sqrt(52)*sqrt(52)/sqrt(52)
    =-6(sqrt(13*2*2) / 52)


    The answer ends up being sin(alpha)= {3(sqrt(13)/13}

    And that is where I am getting lost... It's simpler algebra that I should already know, but I just don't remember. Here's an image that might be easier on the eyes...



    The box is around the answer I should get to. Thanks for any help.
    Attached Thumbnails Attached Thumbnails Trig Help-trig.jpg  
    Last edited by Math Help; September 10th 2005 at 02:15 PM.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    1,631
    You're good.

    Let us continue,
    sin(alpha) = (-6/sqrt(52))
    Rationalize the denominator,
    = (-6/sqrt(52)*sqrt(52)/sqrt(52)
    = -6(sqrt(13*2*2) / 52)
    Get the 2*2 out of the sqrt sign,
    = -6[2sqrt(13)] / 52
    = -12[sqrt(13)] / 52
    Reduce it to its lowest term, divide both numerator and denominator by 4,
    = -3[sqrt(13)] / 13 ----------answer.
    Last edited by ticbol; September 10th 2005 at 01:58 PM.
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  3. #3
    Newbie
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    Sep 2005
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    Thank you very much! Your help is much appreciated
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