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Math Help - [SOLVED] law of cosines issue

  1. #1
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    [SOLVED] law of cosines issue

    I have an issue solving the following problem. This starts using the law of cosines. Anyway, given the following, solve the triangle

    side a = 12
    b = 5
    c = 13

    law of cosines is b^2 = a^2 + c^2 - 2accosB which yields me an angle B of 23 degrees, so far so good.

    so next you're supposed to use the law of sines to solve for the next angle. I'm solving for A for the hell of it, so I get
    \frac{sin23^\circ}{5} = \frac{sinA}{12}
    or
    \frac{12sin23^\circ}{5} = sinA
    that gives me A = 69degrees...but according to the book, it's 67 degrees.

    and if I solve for C instead, I get
    \frac{sin23^\circ}{5} = \frac{sinC}{13}
    or
    \frac{13sin23^\circ}{5} = sin C
    and that doesn't even work, since the left hand of the equation is great than 1, it's out of domain or whatever.

    So...what am I doing wrong?
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  2. #2
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    Do not round (or round to 4 decimal places) until you get to the end. This is where your error is coming from.
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  3. #3
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    Quote Originally Posted by ANDS! View Post
    Do not round (or round to 4 decimal places) until you get to the end. This is where your error is coming from.
    Thanks a lot, I appreciate it. I'm thrilled it's that simple.
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