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Math Help - Find the ANGLE

  1. #16
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    Zewail University - Cairo - Egypt
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    Re: Find the ANGLE

    Quote Originally Posted by razemsoft21 View Post
    Thanks a lot for the steps (logically true)

    but this 0.01 changes 1 degree to 20 degrees
    and changes angle BAC from 99 degrees to 80 degrees.

    again thank you for the logical method.
    hmmm i inderstand thnQ i'm noob in maths
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  2. #17
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    Jerusalem - Israel
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    Re: Find the ANGLE

    Quote Originally Posted by mido22 View Post
    hmmm i inderstand thnQ i'm noob in maths
    sorry, I don't say that.

    a good help from cairo to Palestine (west Bank).
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  3. #18
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    Clarksville, ARk
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    Re: Find the ANGLE

    I get angle BAC = 80 , without rounding anything.
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  4. #19
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    Re: Find the ANGLE

    Hi razemsoft 21,
    AD = 10 sin 70/sin 100 =9.54
    AC = rad3 - 9.54 = 7.78
    cos 50 = AD^2 + DC^2 -AC^2/ 2*AD*DC DC only unknown
    quadratic equation gives two solutions 3,47 is shortest which produces an oblique angle at ACD
    cos DAC = AD^2 +AC^2 -DC^2/2*AD*AC = 20
    ACD =110 BAC= 80

    The sine law could have been used directly to give a value for ACD 0f 70 but since the question wanted an oblique angle the answer would be its supplement.See ambiguous case sin law
    Last edited by bjhopper; August 14th 2011 at 03:09 AM. Reason: correct error
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  5. #20
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    Zewail University - Cairo - Egypt
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    Re: Find the ANGLE

    Quote Originally Posted by bjhopper View Post
    Hi razemsoft 21,
    AD = 10 sin 70/sin 100 =9.54
    AC = rad3 - 9.54 = 7.78
    cos 50 = AD^2 + DC^2 -AC^2/ 2*AD*DC DC only unknown
    quadratic equation gives two solutions 3,47 is shortest which produces an oblique angle at ACD
    cos DAC = AD^2 +AC^2 -DC^2/2*AD*AC = 20
    ACD =110 BAC= 80

    The sine law could have been used directly to give a value for ACD 0f 70 but since the question wanted an oblique angle the answer would be its supplement.See ambiguous case sin law
    good proof i love it
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