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Math Help - Ambigious case for the law of sines?

  1. #1
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    Ambigious case for the law of sines?

    Determine the number of solution for each possible triangle

    B= 61 degrees a=12 b=8
    The answer on the back of my book states zero.
    Although I would think it has 2 because
    12> b sin a when a is 29 degrees
    Can anyone help me as I am slightly confused.
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  2. #2
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    Quote Originally Posted by homeylova223 View Post
    Determine the number of solution for each possible triangle

    B= 61 degrees a=12 b=8
    The answer on the back of my book states zero.
    Although I would think it has 2 because
    12> b sin a when a is 29 degrees
    Can anyone help me as I am slightly confused.
    The triangle must obey the Sine Law, however..

    \displaystyle\frac{sin61^o}{8}\ \ne\ \frac{sin29^o}{12}

    \displaystyle\frac{sinA}{a}=\frac{sinB}{b}\Rightar  row\ sinA=\frac{12sin61^o}{8}=1.312

    which is not possible, since

    -1\ \le\ sinA\ \le\ 1
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  3. #3
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    Wikipedia offers and explanation and drawing which is helpful.

    Law of sines - Wikipedia, the free encyclopedia
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  4. #4
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    Hello, homeylova223!

    Determine the number of solution for this triangle:
    . . B= 61^o,\;a=12,\; b=8

    The answer on the back of my book states: zero..

    If you had made a sketch, the answer would have been obvious.


    Code:
                  C
                  *
                 /:\
                / : \
          a=12 /  :  \ b=8
              /   :   \
             /    :    \
            /     :
           /61o   :
        B * - - - + - - - * A
                  D

    Draw altitude \,CD to side \,AB.
    In right triangle CDB\!:\;\sin61^o \:=\:\dfrac{CD}{12} \quad\Rightarrow\quad CD \:=\:12\sin61^o \;\approx\;10.5

    Side \,b is only 8 units long.

    There is no triangle.

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