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Math Help - given 3 sides, want an angle

  1. #1
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    given 3 sides, want an angle

    i have 3 sides of a nonright triangle , length 5(a), 2(b) and 5.4(c), and i want angle C.
    i have tried law of cosines over and over but keep getting anle of about 90deg, i know it cant be 90deg, because its not a right triangle.
    i must be doing sumthing really stupid, but cant seem to firgue out what is wrong.
    please help somem1?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by helpmewithtrig
    i have 3 sides of a nonright triangle , length 5(a), 2(b) and 5.4(c), and i want angle C.
    i have tried law of cosines over and over but keep getting anle of about 90deg, i know it cant be 90deg, because its not a right triangle.
    i must be doing sumthing really stupid, but cant seem to firgue out what is wrong.
    please help somem1?
    The Law of Cosines ought to work. Perhaps you could write down what it is? Maybe you've got a typo in the equation.

    -Dan
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    what i did

    thanks for the reply topsquark.

    the equation i am using is:

    c^2 = a^2 + b^2 - 2ab cos(C)

    i rearrange to get:

    cos(C) = (a^2 + b^2 - c^2) / -2ab

    when i plug in the numbers i get:

    Cos (C) = -0.008

    and then,

    acrccos (-.008) = 90.46deg

    thanks for any insight.

    a = 5
    b = 2
    c = 5.4
    angle C = ???
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  4. #4
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    Quote Originally Posted by helpmewithtrig
    i have 3 sides of a nonright triangle , length 5(a), 2(b) and 5.4(c), and i want angle C.
    i have tried law of cosines over and over but keep getting anle of about 90deg, i know it cant be 90deg, because its not a right triangle.
    i must be doing sumthing really stupid, but cant seem to firgue out what is wrong.
    please help somem1?
    I think it is easier to use Heron's Theorem for these problems. Do you what it is?
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    herons theorum, isn't that an area formula? thing is, it may be easier, but the cosine law should work right? but i always get 90deg as answer.
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  6. #6
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    Quote Originally Posted by helpmewithtrig
    herons theorum, isn't that an area formula? thing is, it may be easier, but the cosine law should work right? but i always get 90deg as answer.
    The cosine thing would also work just as Heron's Formula use whatever you like more.

    It looks like 90 degress.
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  7. #7
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    cosine law

    any1 know why cosine law wont work???
    thanks some1 out there.
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  8. #8
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    Quote Originally Posted by helpmewithtrig
    any1 know why cosine law wont work???
    thanks some1 out there.
    It always does
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  9. #9
    Forum Admin topsquark's Avatar
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    I'm home now, so I finally have my precious calculator in my hands.

    helpmewithtrig I'm not sure what your problem is. You did the problem correctly. You DON'T have a right triangle because angle C is not 90 degrees, but 90.4584 degrees. There's nothing wrong with this answer!

    -Dan
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  10. #10
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    Quote Originally Posted by topsquark
    I'm home now, so I finally have my precious calculator in my hands.

    helpmewithtrig I'm not sure what your problem is. You did the problem correctly. You DON'T have a right triangle because angle C is not 90 degrees, but 90.4584 degrees. There's nothing wrong with this answer!

    -Dan
    math is the science (or art) of precision
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  11. #11
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    Hi:

    If c^2 = a^2 + b^2 - 2ab cos(C), then cos(C) = (a^2 + b^2 - c^2)/(2ab). You divided by -2ab rather than its positive counterpart. Appropriate substitution of side lengths puts m<(C) at 90.458 to three places.

    Regards,

    Rich B.
    PS: m<(C) translates to "measure of..."
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