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Law of sines Problem

  1. Aug 20, 2007 #1
    Law of sines Problem....

    1. The problem statement, all variables and given/known data
    Solve for angle A:

    sin(135)/56.6 = sin(A)/45 = sin(15)/15



    3. The attempt at a solution

    sin(135)/56.6 = sin(A) = 45sin(15)/15

    Am I on the correct track?
     
  2. jcsd
  3. Aug 20, 2007 #2

    G01

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    You are on the right track, and you have the right idea, but you seem to be confused with that second equal sign.

    [tex]\frac{\sin 135}{56.6}=\frac{\sin A}{45}=\frac{\sin 15}{15}[/tex]

    All this above line says is:

    [tex]\frac{\sin A}{45}= \frac{\sin 15}{15}[/tex]

    and

    [tex]\frac{\sin A}{45}=\frac{\sin 135}{56.6}[/tex]


    All you have to do is pick one of the above equations. They should give you the same answer. Does this help?
     
    Last edited: Aug 20, 2007
  4. Aug 20, 2007 #3
    it sure does... I was trying to solve all together at once. Could not figure out how but that makes sense what you said. Thanks.
     
  5. Aug 20, 2007 #4

    G01

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    No Problem! :smile:
     
  6. Aug 20, 2007 #5

    Dick

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    I'm confused. Why not just use 135+A+15=180 to solve for A? And with the given figures all of the ratios are only roughly equal. Are we really solving the correct problem?
     
  7. Aug 20, 2007 #6

    mgb_phys

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    I suspect we are too clever for our own good!
     
  8. Aug 20, 2007 #7

    HallsofIvy

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    This is the second question I have seen posted under "physics" that had no physics in it! Why isn't this posted under a mathematics thread?
     
  9. Aug 20, 2007 #8

    mgb_phys

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    Because us mere physicists are scared of going over there?
     
  10. Aug 24, 2007 #9
    The question didn't mention that all the angles refer to the actual angles in the triangle. if the question says sin (165)/15. the below method will not work but the above method will work. right?

    135+A+15=180
     
  11. Aug 24, 2007 #10
    if it is not in the right place, why don't the moderator move it?
     
  12. Aug 24, 2007 #11

    Dick

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    If the angles aren't the 'actual angles' of a triangle then what might they be and what could the law of sines have to do with them?
     
  13. Aug 24, 2007 #12
    we must analyze based on what we have
     
  14. Aug 24, 2007 #13

    Dick

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    I give up.
     
  15. Aug 24, 2007 #14
    in case you don't, i will....
     
  16. Aug 24, 2007 #15

    mgb_phys

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    The law of sines only works for the internal angles of a triangle in a flat plane.
    I checked in case it was a trick question about spherical trig.
     
  17. Aug 24, 2007 #16
    They can't be actual angles. They don't satisfy the triangle inequality.
     
  18. Aug 24, 2007 #17
    i been reviewing Trig and i still have yet to finish a chapter that i've been on for 3 days!!! i didn't think Trig would take this much work.
     
  19. Aug 24, 2007 #18

    Dick

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    Then what are they? This is silly. The OP gives more information than you actually need to solve the triangle and that extra information is inconsistant. You can pick three of the given sides or angles and solve a triangle - but you get somewhat different triangles depending on which three you pick.
     
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