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Circle geometry + trig

  1. Nov 5, 2007 #1
    In one of my books there is a question:

    "Problem: What is the trigonometric relationship between the length of the chord and the angle subtended at the centre?"
    [tex]\theta[/tex] is the angle subtended at the centre.

    Next to it is simply written:

    [tex]2r \sin(\frac{\theta}{2}) = l[/tex]

    I'm sure I wrote that; but I can't remember proving it.

    Can anyone help?
    Last edited: Nov 5, 2007
  2. jcsd
  3. Nov 5, 2007 #2

    Gib Z

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    Welcome to PF Jungry !

    Now I'll assume theta is in radians, because the equation is not correct otherwise. However, that equation is true for l being the length of the arc, not the chord :( .

    Try to apply the definition of radian angle measure, and the simple fact that the circumference of 2*pi*r.
  4. Nov 5, 2007 #3
    thanks for the reply.

    Jungry :D

    Lemme draw it up: (click to enlarge I guess)

    http://img451.imageshack.us/img451/4598/probbh2.jpg [Broken]

    Had to cut and paste but :)

    The 2 equations on the left; I think I just wrote them down, but I don't remember how I got there..

    Edit: Now I see my error in the previous post.. try again?
    Last edited by a moderator: May 3, 2017
  5. Nov 5, 2007 #4

    Gib Z

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    Whoops! My bad, Jungy =P

    From the diagram, damn it seems like you did mean chord and not arc, which means the equation you wrote in your first post isn't correct :( Though on the paper you wrote a different thing (which a sine in front of the theta divided by 2). Unfortunately, thats not correct either :(

    Do you perhaps know how to use the Cosine Rule? That is essential here =]
  6. Nov 5, 2007 #5
    [tex]a^2 = b^2+c^2-2bc \cos A[/tex]


    Yeah, I pay soooooo much attention in class :)
    Last edited: Nov 5, 2007
  7. Nov 5, 2007 #6

    Gib Z

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    Very good, you're picking up the [tex]LaTeX[/tex] very fast! Just edit your post and put a space between the \cos and the A, and thats correct =] Use that rule on that triangle in your diagram and the answer comes easily!
  8. Nov 5, 2007 #7
    Yea.. Latex :O

    Edit: Wait..

    [tex] \cos \theta = \frac {b^2 + c^2 - a^2|{2bc} [/tex]

    then sub:

    [tex] \cos \theta = \frac {2r^2 - l^2}{2r^2} [/tex]

    which means [tex] \cos \theta \times 2r^2 = 2r^2 - l^2 [/tex]

    which isnt' getting me anywhere.
    Last edited: Nov 5, 2007
  9. Nov 5, 2007 #8

    Gib Z

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    You should be paying for attention in class if you think [tex]2r^2 - 2r^2\cos \theta = \cos \theta[/tex]!! If I was your teacher I would hit ! *slap!* Think!
  10. Nov 5, 2007 #9
    Following on from my previous post..

    [tex] l^2 = 2r^2 (1 - \cos \theta) [/tex]


    Which brings us to:

    [tex] \frac {l^2}{(1 - \cos \theta)} = 2r^2 [/tex]


    Wait I need to make l the subject..
  11. Nov 5, 2007 #10

    Gib Z

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    That is correct =]
  12. Nov 5, 2007 #11
    Then what the hell does [tex] 2r \sin (\frac { \theta}{2}) = l [/tex] have to do with this T_T..
  13. Nov 5, 2007 #12

    Gib Z

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    I really don't know >.< Perhaps you copied down the wrong thing?
  14. Nov 5, 2007 #13
    Well I'm pretty sure my teacher gave that to us...


    -scribble scribble-

    Probably why I don't pay attention in class :D

    And for new questions do I have to start a new topic, or can I just continue rambling on in here?
  15. Nov 5, 2007 #14

    Gib Z

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    If they're on the same subject, ie Circle Geo and trig, then I guess its fine. otherwise just start a new thread.
  16. Nov 5, 2007 #15
    Alright I'll start a new one later.

    Thanks for the help Gib Z.

    I hope to the Lord that's not in my exam tomorrow.
  17. Nov 5, 2007 #16

    Gib Z

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    O just before you start the new thread, make sure its in the Pre Calc Homework section instead of general math. Bye for now then. Good luck on the exam.
  18. Nov 5, 2007 #17
    Cool thanks.
  19. Dec 11, 2007 #18
    Why do you have to use the cosing rule here? He has a mistake in his attachment in the has sin(th/2) expression on the right side. Just apply sine(th/2) and you'll get the answer. No?
  20. Dec 11, 2007 #19

    Ben Niehoff

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    This formula is correct! I don't know why GibZ is telling you otherwise.

    The chord is the straight line. If you bisect the angle, you will create two right triangles each with hypotenuse r and side L/2. That's where the formula comes from. You don't need the law of cosines for anything.
  21. Dec 12, 2007 #20

    Gib Z

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    It turns out that formula is actually equivalent to the same form I use, though I did not recognize it. I personally derive it with the law of Cosines as thus;

    A chord is a length l, subtended from the center of the circle with radius r, at theta radians.

    By the law of cosines:
    [tex]l^2 = r^2 + r^2 - 2\cdot r \cdot r \cdot \cos \theta = 2r^2 (1 - \cos \theta)[/tex]

    That is the form I usually leave it at, though at hindsight I should have seen the simple manipulation;
    [tex]l^2 = 2r^2 (1 - \cos \theta) = 4r^2 \frac{ (1 - \cos \theta) }{2} = 4r^2 \sin^2 (\theta /2)[/tex], and since the length must be positive, [tex] l = 2r \sin (\theta /2)[/tex]

    I do have a point of defense though =] At least the OP has a method that 1) is still a correct and viable method and 2) knows how to derive the result, rather than remember one he copied from his lecture.
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