# Circle geometry + trig

by Jungy
Tags: circle, geometry, trig
 P: 27 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?" $$\theta$$ is the angle subtended at the centre. Next to it is simply written: $$2r \sin(\frac{\theta}{2}) = l$$ I'm sure I wrote that; but I can't remember proving it. Can anyone help?
 HW Helper P: 3,348 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.
 P: 27 thanks for the reply. Jungry :D Lemme draw it up: (click to enlarge I guess) 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?
 HW Helper P: 3,348 Circle geometry + trig 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 =]
 P: 27 $$a^2 = b^2+c^2-2bc \cos A$$ That? Yeah, I pay soooooo much attention in class :)
 HW Helper P: 3,348 Very good, you're picking up the $$LaTeX$$ 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!
 P: 27 Yea.. Latex :O Edit: Wait.. $$\cos \theta = \frac {b^2 + c^2 - a^2|{2bc}$$ then sub: $$\cos \theta = \frac {2r^2 - l^2}{2r^2}$$ which means $$\cos \theta \times 2r^2 = 2r^2 - l^2$$ which isnt' getting me anywhere.
 HW Helper P: 3,348 You should be paying for attention in class if you think $$2r^2 - 2r^2\cos \theta = \cos \theta$$!! If I was your teacher I would hit ! *slap!* Think!
 P: 27 Following on from my previous post.. $$l^2 = 2r^2 (1 - \cos \theta)$$ Right? Which brings us to: $$\frac {l^2}{(1 - \cos \theta)} = 2r^2$$ Edit: Wait I need to make l the subject..
 HW Helper P: 3,348 That is correct =]
 P: 27 Then what the hell does $$2r \sin (\frac { \theta}{2}) = l$$ have to do with this T_T..
 HW Helper P: 3,348 I really don't know >.< Perhaps you copied down the wrong thing?
 P: 27 Well I'm pretty sure my teacher gave that to us... Bah! -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?
 HW Helper P: 3,348 If they're on the same subject, ie Circle Geo and trig, then I guess its fine. otherwise just start a new thread.
 P: 27 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.
 HW Helper P: 3,348 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.
 P: 27 Cool thanks.
 P: 137 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?

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