Circle geometry + trig


by Jungy
Tags: circle, geometry, trig
Jungy
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#1
Nov5-07, 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?
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Gib Z
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#2
Nov5-07, 03:47 AM
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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.
Jungy
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#3
Nov5-07, 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?

Gib Z
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#4
Nov5-07, 04:25 AM
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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 =]
Jungy
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#5
Nov5-07, 04:28 AM
P: 27
[tex]a^2 = b^2+c^2-2bc \cos A[/tex]

That?

Yeah, I pay soooooo much attention in class :)
Gib Z
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#6
Nov5-07, 04:30 AM
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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!
Jungy
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#7
Nov5-07, 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.
Gib Z
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#8
Nov5-07, 04:39 AM
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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!
Jungy
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#9
Nov5-07, 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..
Gib Z
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#10
Nov5-07, 04:46 AM
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That is correct =]
Jungy
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#11
Nov5-07, 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..
Gib Z
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#12
Nov5-07, 04:51 AM
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I really don't know >.< Perhaps you copied down the wrong thing?
Jungy
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#13
Nov5-07, 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?
Gib Z
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#14
Nov5-07, 04:55 AM
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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.
Jungy
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#15
Nov5-07, 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.
Gib Z
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#16
Nov5-07, 05:00 AM
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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.
Jungy
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#17
Nov5-07, 05:09 AM
P: 27
Cool thanks.
unplebeian
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#18
Dec11-07, 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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