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Homework Help: Cos B

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data

    Find Cos B

    Hint: use trigonometry + Geometry

    http://img210.imageshack.us/img210/4596/pow.png [Broken]

    2. Relevant equations

    trigonometry + Geometry

    3. The attempt at a solution

    I dont see how i could use trig to find any of the angles, and since none of the lins go through the center it doesnt help me find the raudius. I dont even know how to begin solving this.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 31, 2010 #2

    tiny-tim

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    Hi um0123! :smile:

    Hint: It's a circle, so B = 180º - D. :wink:
     
  4. Jan 31, 2010 #3
    i may be daft, but may i have a little more help? i see how that information would be really useful but i cant seem to grap what i need to do.
     
  5. Jan 31, 2010 #4

    tiny-tim

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    Hint: AC2 = … ? :smile:
     
  6. Jan 31, 2010 #5
    [tex]AC^2 = 1^2 + 9^2 - 4(1)(9)cos(180-D)[/tex]

    [tex]AC^2 = 82 - 34cos(180-D)[/tex]

    [tex]\frac{AC^2 - 82}{-34} = cos(180-D)[/tex]

    [tex]\frac{AC^2 - 82}{-34} = cos(180)cos(D) + sin(180)sin(D)[/tex]

    [tex]\frac{AC^2 - 82}{-34} = -1cos(D) + 0sin(D)[/tex]


    [tex]\frac{AC^2 - 82}{-34} = -1cos(D)[/tex]


    [tex]\frac{AC^2 - 82}{34} = cos(D)[/tex]

    now im stuck
     
  7. Jan 31, 2010 #6

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    erm :redface: … AC2 = 62 + 92 - … ? :smile:
     
  8. Jan 31, 2010 #7
    i cant assume its a right triangle, can i? i was using the law of cosines because the Pythagorean theorem is only for right triangles, unless i can prove its a right triangle i don't think im allowed to use that.

    BTW what does the X^2 tag do if i put it in the tag box?
     
  9. Jan 31, 2010 #8

    Mark44

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    No, it's not a right triangle. tiny-tim was not saying you shouldn't use the Law of Cosines, but he was saying that you should use it correctly, and was trying to guide you in that direction.
     
  10. Jan 31, 2010 #9

    tiny-tim

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    I meant there are two triangles with AC …

    you've only used one of them!
     
  11. Jan 31, 2010 #10
    oh, sorry, i didn't notice the minus sign after your previous post tiny-tim (my eyesight is less than acceptable).

    so i continue with:


    [tex]AC^2 = 6^2 + 9^2 - 4(6)(9)cos(D) [/tex]

    [tex]AC^2 = 36 + 81 - 216cos(D) [/tex]

    [tex]AC^2 = 117 - 216cos(D) [/tex]

    [tex]AC^2 -117 = -216cos(D) [/tex]

    [tex]\frac{AC^2 -117}{-216} = cos(D) [/tex]

    but i still get stuck....i must be overlooking something really obvious.
     
    Last edited: Jan 31, 2010
  12. Jan 31, 2010 #11

    tiny-tim

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    Forget the fractions …

    you now have two expressions for AC2 (one for each triangle),

    so put them equal, and that gives you an equation in cosD. :smile:
     
  13. Jan 31, 2010 #12
    You can use the cosine rule in the triangle ADC as well, giving another expression for AC and cos(D).
    You put a 4 in the cosine rule where a 2 was wanted
     
  14. Jan 31, 2010 #13
    So if i understand you correctly, tiny-tim, i should have:
    [tex] 117 - 216Cos(D) = 82 - 34Cos(180-D) [/tex]

    [tex] 35 - 216Cos(D) = -34Cos(180-D) [/tex]

    [tex] \frac{35 - 216Cos(D)}{-34} = -1Cos(180-D) [/tex]

    [tex] \frac{35 - 216Cos(D)}{34} = Cos(180-D) [/tex]

    [tex] \frac{35 - 216Cos(D)}{34} = Cos(180)Cos(D) + Sin(180)Sin(D) [/tex]

    [tex] \frac{35 - 216Cos(D)}{34} = -1Cos(D) + 0Sin(D) [/tex]

    [tex] \frac{35}{34} = \frac{216Cos(D)}{34} - Cos(D)[/tex]

    now i really cant go anywhere
     
    Last edited: Jan 31, 2010
  15. Jan 31, 2010 #14

    tiny-tim

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    (I haven't checked the arithmetic, but …)

    that should give you an equation of the form PcosD = Q.
     
  16. Jan 31, 2010 #15
    im sorry, i dont think i have learned that equation, if i have then i am not recognizing it.

    btw, i just wanted to say thanks for helping me through this, i know im probably really hard to handle.
     
  17. Jan 31, 2010 #16
    You're almost there. You just have to give the 2 fractions on the right the same denominator so you can add them.

    You also need to use the correct cosine rule: a^2 = b^2 + c^2 -2 bc cos(A)
     
  18. Jan 31, 2010 #17
    oh, wow, i thought it was 4bc cos(a)

    also i did a mathematical error when i calculated 4(1)(9) as 34, it sohuld be 36,

    so redoing the calculations i get:

    [tex] \frac{35}{18} = \frac{108cos(D)}{18} - cos(D) [/tex]

    [tex] \frac{35}{18} = 6cos(D) - cos(D)[/tex]

    [tex] \frac{35}{18} = 5cos(D) [/tex]

    [tex]35 = 90cos(D) [/tex]

    [tex]\frac{35}{90} = cos(D) [/tex]

    please tell me im right!?
     
    Last edited: Jan 31, 2010
  19. Jan 31, 2010 #18
    form there all i have to do is cos^-1(35/90) to ge tthe angle and subtract from 180. I hope this is correct.


    Thanks so much for you help, tiny-tim and willem2!
     
  20. Feb 1, 2010 #19
    Unfortunately there's a sign error here, that's still present in your final answer.
     
  21. Feb 1, 2010 #20
    Express AC² in terms of cosB and cosD.
    Now recall cosD=cos(180-B)=-cosB
    Try to get cos B
     
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