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Given isosceles triangle, find sin (A-C)

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  1. Aug 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Triangle ABC have side AB = 10 cm, AC=BC = 13cm, so sin (A-C) is...

    2. Relevant Equations

    sin (A-C) = sin A cos C - cos A sin C

    3. The attempt at a solution

    I see that the triangle can be split into two right-angle triangles.
    But, sin (A-C) ?? How to get that?
     
  2. jcsd
  3. Aug 24, 2015 #2

    SteamKing

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    Is there a sketch of this triangle?
     
  4. Aug 24, 2015 #3
    Actually, there is no sketch of the triangle in the question.
    But, here is my sketch
    xzoifdlohaoniigd.png
     
  5. Aug 24, 2015 #4
    use sohcahtoa

    you know all the lengths, you can easily now work out all the angles.

    http://www.mathwords.com/s/s_assets/s126.gif
     
  6. Aug 24, 2015 #5
    Where did that equation come from?
     
  7. Aug 24, 2015 #6

    SammyS

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    It's one of the well known angle sum/difference formulas.
     
  8. Aug 24, 2015 #7
    True, but does he need that here?
     
  9. Aug 24, 2015 #8

    SammyS

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    Is it necessary? I don't know.

    However, it can be used to solve the problem
     
  10. Aug 24, 2015 #9
    Seems like the wrong equation to apply to me.

    Edit: Never mind. Now I see. The next step makes it a simple multiplication, division and subtraction problem. That is a very elegant solution when applied that way. Cause we can apply the lengths directly !

    We use the equations WW posted, but rather than calculate the angles, we eliminate the Sin's and Cos's. to get a simple math expression! No trig tables or Scientific calculator needed.

     
    Last edited: Aug 24, 2015
  11. Aug 25, 2015 #10
    So, should I use the formula sin (A-C) = sin A cos C - cos A sin C ??

    sin A = 12/13
    cos A = 5/13

    sin (C/2) = 5/13
    cos (C/2) = 12/13

    Then,
    sin (C) = 2 * sin (C/2) * cos (C/2) = 2 * (5/13) * (12/13) = 120/169
    cos (C) = 1 - 2 sin^2(C/2) = 1 - 2 * 25/169 = 1 - 50/169 = 119/169

    So,

    sin (A - C) = sin A cos C - cos A sin C = 12/13 * 119/169 - 5/13*120/169 = 1428/2197 - 600/2197 = 828/2197

    Do I get it right ?? Is there any simpler solution ?
     
  12. Aug 25, 2015 #11
    Besides using a calculator with trig functions?
    A = asin 12/13.
    B=A
    C = 180 - A - B

    Dif = A-C

    answer = Sin (dif)

    edited. And edited
    Sin (A) = (12/13) = 0.923. A = Asin (1.17) = 67.38 degrees. Same as B. C= 180-2A = 45.23
    A-C = 67.38 - 45.23 = 22.14
    Sin (22.14degrees) = sin (0.386 radians) = 0.376878. yes, your answer appears to be correct !
     
    Last edited: Aug 25, 2015
  13. Aug 25, 2015 #12
    By golly, it appears that you got it right ! (Took me a while to get the calculation right the other way)
     
    Last edited: Aug 25, 2015
  14. Aug 25, 2015 #13
    Yap, using calculator is the simplest way, but it's not allowed in the test :(
     
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