What is the Trigonometric Relationship Between Chord Length and Angle at Centre?

[Jungy]

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?

 

[Gib Z]

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]

thanks for the reply.

Jungry :D

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

http://img451.imageshack.us/img451/4598/probbh2.jpg

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]

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, that's not correct either :(

Do you perhaps know how to use the Cosine Rule? That is essential here =]

 

[Jungy]

$$a^2 = b^2+c^2-2bc \cos A$$

That?

Yeah, I pay soooooo much attention in class :)

 

[Gib Z]

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 that's correct =] Use that rule on that triangle in your diagram and the answer comes easily!

 

[Jungy]

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.

 

[Gib Z]

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!

 

[Jungy]

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..

 

[Gib Z]

That is correct =]

 

[Jungy]

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

 

[Gib Z]

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

 

[Jungy]

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]

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

 

[Jungy]

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]

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]

Cool thanks.

 

[unplebeian]

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?

 

[Ben Niehoff]

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:

$$2r \sin(\frac{\theta}{2}) = l$$

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.

 

[Gib Z]

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:
$$l^2 = r^2 + r^2 - 2\cdot r \cdot r \cdot \cos \theta = 2r^2 (1 - \cos \theta)$$

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

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.

 

[Ben Niehoff]

I suspected your formula might be equivalent, but I was too lazy to apply trig identities. :P

My excuse is that I'm spending all my time writing applications for grad school. Unfortunately, that has to be done on the computer...and so I can't completely escape the distraction of PF. :P