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Math Help - An SAS triangle can't have two solutions... but this one is strange

  1. #1
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    An SAS triangle can't have two solutions... but this one is strange

    <A = 34 degrees. b=24. c=46.

    This is an SAS triangle... when I use Law of Cosines, I solved for side a and got a=29.35.

    Then I tried using Law of Sines to solve for <B and <C.

    If I solve for <B first, I set it up this way:

    sin(34)
    -------
    29.35

    =

    sin B
    --------
    24

    and I get <B=27.2 degrees. Which means <C=118.8 degrees.

    However, if I solve for <C first, I set it up this way:

    sin(34)
    -------
    29.35

    =

    sin C
    --------
    46

    and I get <C=61.2 degrees. Which means <B=84.8 degrees.


    What is going on?! I know I'm missing something here, but I can't figure out what it is!

    Please help!
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  2. #2
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    Quote Originally Posted by Mpromptu View Post
    <A = 34 degrees. b=24. c=46.

    a=29.35. OK

    Then I tried using Law of Sines to solve for <B and <C.

    <C=118.8 degrees. OK


    However, if I solve for <C first, I set it up this way:

    sin(34)
    -------
    29.35

    =

    sin C
    --------
    46

    and I get <C=61.2 degrees.
    1. If you consider the lengths of the sides of your triangle you'll see that the angle C must be very much larger than the other two angles.

    2. From

    \frac{\sin(34^\circ)}{29.35} = \frac{\sin(C)}{46} ~\implies~\sin(C)=0.876418...

    And with this value you get two different angles for C:

    C\approx 61.2^\circ~\vee~ C=180^\circ-61.2^\circ=118.8^\circ
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  3. #3
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    Thanks for the reply, but I don't think my question has been answered.

    Obviously, <C must be the largest of my two choices; it's opposite the largest side. Choosing the smaller <C would make no sense, visually.

    My question is why I have two choices for <C to begin with! This is an SAS triangle. Shouldn't there only be one option for my missing information?

    For what reasons can I dismiss the smaller C? Is there any mathematical reason I'm even coming up with that answer in the first place?
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  4. #4
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    Quote Originally Posted by Mpromptu View Post
    Thanks for the reply, but I don't think my question has been answered.

    Obviously, <C must be the largest of my two choices; it's opposite the largest side. Choosing the smaller <C would make no sense, visually.

    My question is why I have two choices for <C to begin with! This is an SAS triangle. Shouldn't there only be one option for my missing information?

    For what reasons can I dismiss the smaller C? Is there any mathematical reason I'm even coming up with that answer in the first place?
    Let us pretend \angle C = 61.2^\circ

    With \angle A = 34^\circ you'll get \angle B = 84.8^\circ

    That means \angle B is the greatest angle but b is the smallest side, which is impossible. Therefore \angle C = 61.2^\circ must be false.
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  5. #5
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    It's because your calculator is picking it up in the first quadrant rather than the second. If you remember your good ole unit circle, you understand the sine of say. . . 60 degrees will be the same as the sine of 120 degrees. When you take that inverse sine in the end it is giving you a right answer, just in the wrong quadrant.
    Check this out: 90-61.2=28.8
    118.8-90=28.8

    Do you dig that? I know the best way to avoid this with an SSS triangle is to always solve for the biggest angle first. I hope you sort of understand now and can get around this in the future.
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  6. #6
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    Thanks, thetreece

    I think that's the issue. I appreciate the help!
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