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Math Help - trig help

  1. #1
    Senior Member
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    trig help

    I have to solve a triangle

    the given information in triangle ABC is c=28.6,b=19.9 and A=48.1^o

    i used the cosine law to get a which i found to be 21.3 is this correct?

    i then tried to use the cosine law again to find one of the other angles

    my work:

    <br />
c^2=a^2+b^2=2ab CosC

    <br />
28^2=21.3^2+19.9^2-2(21.3)(19.9)CosC

    <br />
CosC=\frac{28.6^2-21.3^2-19.9^2}{-2(21.3)(19.9)}

    <br />
CosC= \frac{-31.74}{-822.18}

    CosC=sin^{-1} 0.038

    C= 87^o

    is this right? i don't think it is because we are working with obtuse triangles
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  2. #2
    Junior Member
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    Correct, and keep in mind that the law of cosines applies to triangles in general, not just obtuse ones.
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  3. #3
    Member TheMasterMind's Avatar
    Joined
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    Quote Originally Posted by william View Post
    I have to solve a triangle

    the given information in triangle ABC is c=28.6,b=19.9 and A=48.1^o

    i used the cosine law to get a which i found to be 21.3 is this correct?

    i then tried to use the cosine law again to find one of the other angles

    my work:

    <br />
c^2=a^2+b^2=2ab CosC

    <br />
28^2=21.3^2+19.9^2-2(21.3)(19.9)CosC

    <br />
CosC=\frac{28.6^2-21.3^2-19.9^2}{-2(21.3)(19.9)}

    <br />
CosC= \frac{-31.74}{-822.18}

    CosC=sin^{-1} 0.038

    C= 87^o

    is this right? i don't think it is because we are working with obtuse triangles
    The work you have done appears correct. Now you have the 87^o and 48.1^o you can simply find the last angle and you're done.
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