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Math Help - Law of sine and cosine problem

  1. #1
    jwu
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    Law of sine and cosine problem

    Just a simple problem. But I got stuck on some points.
    We just learned the law of sine and the law of cosine from class.
    And here's the assignment according this two topics:

    Solve each triangle. Give lengths to three significant digits and angle measures to the nearest tenth of a degree.
    a=8, b=5, ∠C=60

    My solution:
    Because we have the condition "SAS". So I used the law of cosine to solve for c.
    c=a+b-2abcos 60
    =7

    Then all we need to do next is to find the two angles of the triangle, right?
    So I used the sine law:
    sin C/c=sin A/a
    sin 60/7=sin A/8
    sin A= sin 60*8/7
    A= 81.8 and 98.2

    Here's where I got stuck. The answer at the very end of book only shows the answer 81.8. But I think 98.2 is also acceptable. There should be two possiblities. Can somebody tell me why it doesn't fit?
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by jwu View Post
    Just a simple problem. But I got stuck on some points.
    We just learned the law of sine and the law of cosine from class.
    And here's the assignment according this two topics:

    Solve each triangle. Give lengths to three significant digits and angle measures to the nearest tenth of a degree.
    a=8, b=5, ∠C=60

    My solution:
    Because we have the condition "SAS". So I used the law of cosine to solve for c.
    c=a+b-2abcos 60
    =7

    Then all we need to do next is to find the two angles of the triangle, right?
    So I used the sine law:
    sin C/c=sin A/a
    sin 60/7=sin A/8
    sin A= sin 60*8/7
    A= 81.8 and 98.2

    Here's where I got stuck. The answer at the very end of book only shows the answer 81.8. But I think 98.2 is also acceptable. There should be two possiblities. Can somebody tell me why it doesn't fit?
    Hi jwu,

    You were initially given two sides and an included angle.
    You used the law of cosine to find side C.

    So, now you have 3 fixed lengths and one angle.

    The ambiguous case comes into play when you are given
    two sides and an angle opposite one of them.

    This is obviously not the case here.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by jwu View Post
    Just a simple problem. But I got stuck on some points.
    We just learned the law of sine and the law of cosine from class.
    And here's the assignment according this two topics:

    Solve each triangle. Give lengths to three significant digits and angle measures to the nearest tenth of a degree.
    a=8, b=5, ∠C=60

    My solution:
    Because we have the condition "SAS". So I used the law of cosine to solve for c.
    c=a+b-2abcos 60
    =7

    Then all we need to do next is to find the two angles of the triangle, right?
    So I used the sine law:
    sin C/c=sin A/a
    sin 60/7=sin A/8
    sin A= sin 60*8/7
    A= 81.8 and 98.2

    Here's where I got stuck. The answer at the very end of book only shows the answer 81.8. But I think 98.2 is also acceptable. There should be two possiblities. Can somebody tell me why it doesn't fit?
    Hi jwu,

    another way to look at this is...

    if you draw the diagram, evenly roughly will do,
    notice that the triangle's shape is fully determined
    after you've drawn the 2 sides and the included angle.

    All that's left is to join the remaining two endpoints.
    There's no other way to draw this triangle.

    Hence the remaining angles are already fixed.

    You can verify side "a" with the law of cosines.

    a^2=b^2+c^2-2bccosA

    8^2=5^2+7^2-2(5)7cosA

    64=25+49-70cosA

    70cosA=74-64=10

    cosA=\frac{1}{7}

    A=cos^{-1}\left(\frac{1}{7}\right)=81.78^o

    or 360^o-81.78^o!! which we can certainly discard!
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