
#1
Nov507, 01:38 AM

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?" [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? 



#2
Nov507, 03:47 AM

HW Helper
P: 3,353

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. 



#3
Nov507, 04:21 AM

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? 



#4
Nov507, 04:25 AM

HW Helper
P: 3,353

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 =] 



#5
Nov507, 04:28 AM

P: 27

[tex]a^2 = b^2+c^22bc \cos A[/tex]
That? Yeah, I pay soooooo much attention in class :) 



#6
Nov507, 04:30 AM

HW Helper
P: 3,353

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!




#7
Nov507, 04:32 AM

P: 27

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. 



#8
Nov507, 04:39 AM

HW Helper
P: 3,353

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!




#9
Nov507, 04:44 AM

P: 27

Following on from my previous post..
[tex] l^2 = 2r^2 (1  \cos \theta) [/tex] Right? Which brings us to: [tex] \frac {l^2}{(1  \cos \theta)} = 2r^2 [/tex] Edit: Wait I need to make l the subject.. 



#11
Nov507, 04:48 AM

P: 27

Then what the hell does [tex] 2r \sin (\frac { \theta}{2}) = l [/tex] have to do with this T_T..




#12
Nov507, 04:51 AM

HW Helper
P: 3,353

I really don't know >.< Perhaps you copied down the wrong thing?




#13
Nov507, 04:54 AM

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? 



#14
Nov507, 04:55 AM

HW Helper
P: 3,353

If they're on the same subject, ie Circle Geo and trig, then I guess its fine. otherwise just start a new thread.




#15
Nov507, 04:57 AM

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. 



#16
Nov507, 05:00 AM

HW Helper
P: 3,353

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.




#17
Nov507, 05:09 AM

P: 27

Cool thanks.




#18
Dec1107, 12:42 PM

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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