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Math Help - Can someone find the mistake?

  1. #1
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    Can someone find the mistake?

    Solve the triangle:
    b=2 c=3 A=95 (Capital letters are angles, lower case is sides.)

    a=sqrt(2^2+3^2-2(2)(3)cos(95))
    a= 3.7 (Rounded.)

    b^2=a^2+c^2-2(a)(c)cosB
    2(a)(c)cosB=a^2+c^2-b^2
    cosB=[a^2+c^2-b^2]/[2(a)(c)]
    B=arccos([a^2+c^2-b^2]/[2(a)(c)])
    B=arccos([3.7^2+3^2-2^2]/[2(3.7)(3)])=32.66 (Rounded)

    By the same logic then...
    C=arccos([3.7^2+4^2-3^2]/[2(3.7)(2)])=54.04

    However;
    95+32.66+54.04=181.7
    Ergo, something's wrong, but what? I think B might equal 30.96, but I don't know where I went wrong in the math...
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Re: Can someone find the mistake?

    this error happened from the rounding process it is normal nothing wrong
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  3. #3
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    Clarksville, ARk
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    Re: Can someone find the mistake?

    You have a Typo in your expression for B, but your calculation used the correct value.

    I'm pretty sure that the reason for your angle discrepancy is that you rounded the value for a quite severely. The actual value is more like 3.748 .

    That's more than a 1% difference. Your angle error is about 1% also.
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