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

In summary, the conversation discusses a trigonometric relationship between the length of a chord and the angle subtended at the center. A question is posed about proving the relationship and assistance is offered using the law of cosines. The formula 2r sin(theta/2) = l is also mentioned and one participant defends its validity. The conversation ends with a discussion on personal distractions and the use of trig identities.
  • #1
Jungy
27
0
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:
Mathematics news on Phys.org
  • #2
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
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:
  • #4
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 =]
 
  • #5
[tex]a^2 = b^2+c^2-2bc \cos A[/tex]

That?

Yeah, I pay soooooo much attention in class :)
 
Last edited:
  • #6
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 that's correct =] Use that rule on that triangle in your diagram and the answer comes easily!
 
  • #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:
  • #8
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
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..
 
  • #10
That is correct =]
 
  • #11
Then what the hell does [tex] 2r \sin (\frac { \theta}{2}) = l [/tex] have to do with this T_T..
 
  • #12
I really don't know >.< Perhaps you copied down the wrong thing?
 
  • #13
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
If they're on the same subject, ie Circle Geo and trig, then I guess its fine. otherwise just start a new thread.
 
  • #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.
 
  • #16
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
Cool thanks.
 
  • #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?
 
  • #19
Jungy said:
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]

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.
 
  • #20
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.
 
  • #21
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
 

1. What is the formula for finding the circumference of a circle?

The formula for finding the circumference of a circle is C = 2πr, where C represents the circumference and r represents the radius of the circle. This formula is derived from the relationship between the circumference and diameter of a circle, which is C = πd.

2. How do you find the area of a circle using trigonometry?

To find the area of a circle using trigonometry, you can use the formula A = πr^2, where A represents the area and r represents the radius of the circle. This formula is derived from the relationship between the area of a circle and the radius, which is A = ½r^2.

3. What is the relationship between the trigonometric functions and circle geometry?

The trigonometric functions (sine, cosine, tangent) are closely related to circle geometry. In a right triangle, the sine of an angle is equal to the ratio of the opposite side to the hypotenuse, the cosine is equal to the adjacent side to the hypotenuse, and the tangent is equal to the opposite side to the adjacent side. These relationships can also be applied to points on a circle, where the sine, cosine, and tangent of an angle are equal to the y-coordinate, x-coordinate, and slope of a line passing through that angle on the circle, respectively.

4. How do you use trigonometry to find missing angles in a circle?

To find missing angles in a circle using trigonometry, you can use the inverse trigonometric functions (arcsine, arccosine, arctangent). For example, if you know the ratio of the opposite side to the hypotenuse in a right triangle, you can use the arcsine function to find the measure of the angle. Similarly, if you know the slope of a line passing through a point on a circle, you can use the arctangent function to find the measure of the angle.

5. How can you use the Pythagorean theorem to solve problems involving circle geometry and trigonometry?

The Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides, can be used to solve problems involving circle geometry and trigonometry. For example, if you know the length of the radius and the distance between two points on the circle, you can use the Pythagorean theorem to find the length of the chord connecting those two points.

Similar threads

  • General Math
Replies
7
Views
8K
Replies
1
Views
1K
  • General Math
Replies
17
Views
4K
Replies
7
Views
2K
  • General Math
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
740
Replies
4
Views
192
  • DIY Projects
Replies
6
Views
3K
Replies
8
Views
1K
Back
Top