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Can you tell me why my trig functions aren't working?

  1. Sep 11, 2012 #1
    I am given two sides of a triangle and the angle in/between them: 9 in/s and 4.5 in/s at 50 degrees. I am using the Law of cosines to get the third side which is 7.013 in/s. I then used the law of sine to find the two remaining angles. I have continually gotten 79.4 for one angle and 29.4 for the other. 80 + 30 + 50 = 160. I can't figure out what I am doing wrong. Will someone please tell me the correct solutions, it's killing me. I've been attempting this for 1 hour with consistent results. Thanks, I'm a dummy...
     
  2. jcsd
  3. Sep 11, 2012 #2

    phinds

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    the angle that you think is 79.4 is actually 180 MINUS 79.4 and I have no idea where you got 29.4 for anything but once you change the 79.4 maybe you'll get the other one right too.
     
  4. Sep 11, 2012 #3

    rl.bhat

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    Show your calculations.
     
  5. Sep 11, 2012 #4
    Calculations:
    [itеx] v(a)^2 = 9^2 +4.5^2 - 2 * 4.5 * 9 cos (50)
    v(a) = 7.013

    7.013/Sin(50) = 4.5/ Sin(alpha)
    alpha = 29.44

    7.013/sin(50) = 9/sin(beta)
    beta = 79.44
    [/itеx]
     
  6. Sep 11, 2012 #5
    Calculations:
    [itеx] v(a)^2 = 9^2 +4.5^2 - 2 * 4.5 * 9 cos (50)
    v(a) = 7.013

    7.013/Sin(50) = 4.5/ Sin(alpha)
    alpha = 29.44

    7.013/sin(50) = 9/sin(beta)
    beta = 79.44
    [/itеx]
     
  7. Sep 11, 2012 #6

    rl.bhat

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    Angle between 9 and 7.013 is more than 90 degrees. You can verify this by using cosine rule to find the angle between them.
     
  8. Sep 11, 2012 #7
    used the law of cosines. still got 29.44
     
  9. Sep 11, 2012 #8

    rl.bhat

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    Sorry. Use cosine rule to find the angle between 4.5 and 7.013.
     
  10. Sep 11, 2012 #9
  11. Sep 11, 2012 #10

    rl.bhat

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    Is there no negative sign in the answer?
     
  12. Sep 11, 2012 #11
  13. Sep 11, 2012 #12
    the inverse is negative
     
  14. Sep 11, 2012 #13
    The angle beta could be either 79.44 or (180 - 79.44), as determined by the law of sines.
     
    Last edited: Sep 11, 2012
  15. Sep 11, 2012 #14
    what does that mean? I thought this was a law? Do you have to use the same law for all angles?
     
  16. Sep 11, 2012 #15
  17. Sep 11, 2012 #16

    rl.bhat

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    In the second quadrant cosine is negative. So whatever angle you get, you have to write it as 180-θ
     
  18. Sep 11, 2012 #17
    I see.. but when you use the laws to obtain angle B, why do I get 79.44? Makes me doubt the laws. When do you have to subtract the angle from 180. Is there a rule that makes since of this?
     
  19. Sep 11, 2012 #18
    sin(θ) = sin(180 - θ)
     
  20. Sep 11, 2012 #19
    It how the law was derived.
    In this example one of the common perperdicular line(common to adjacent angles) is outside the triangle.
    Thus we are measuring the external angle.

    Thanks for bring it up. I think that it is just plug and chug.

    Add: It's really prove that a diagram or a sketch is very helpful as in tool of problem solving, IDEA(D for drawing) from Richard Wolfson's book.
     
    Last edited: Sep 11, 2012
  21. Sep 12, 2012 #20
    No the laws are correct. There are two possible solutions obtained applying the law of sines to this problem. Both solutions satisfy the law of sines. Applying the law of cosines to the other angles of the triangle clinches the results. In fact, applying the law of cosines delivers a unique solution.
     
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