# Trig question - segment angle

1. Nov 22, 2006

### S_BX

Hi,
I have a segment of a circle, and know the following quantities.
I need to find cos(BETA), and I know that it's supposed to be = (L^2 - 4h^2)/(L^2 + 4h^2), and I've been trying this for hours.. I suppose it has something to do with the triangle - as you can find the 3rd side (=radius-h), but I just can't get it.. Can somebody please help me out?

http://www.stevenbetts.btinternet.co.uk/bangle.gif

Last edited by a moderator: Apr 22, 2017
2. Nov 22, 2006

### lotrgreengrapes7926

You could use the Pythagorean Theorem to find the 3rd side. But, there is too much information, so $$\cos\beta$$ can be expressed many ways.

3. Nov 23, 2006

### S_BX

I just took the 3rd side to be R-h.

4. Nov 23, 2006

### Integral

Staff Emeritus
Could you show me how you came up with

$$\frac {L^2 + 4h^2} {8h}$$ for the hypotenuse. Why not just call it what it is R?

5. Nov 23, 2006

### S_BX

Because the 1st part of the question asked me to show that R = that. I got it by Pythagoras theorem...

R^2 = (L/2)^2 + (R-h)^2 (pythag)
R^2 = L^2/4 + R^2 - 2hR + h^2 (expand square terms)
2hR = L^2/4 + h^2 (gather like terms)
8hR = L^2 + 4h^2 (multiply through by 4)
R = (L^2 + 4h^2)/8h (rearrange for R)

The second part of the question is to find cos(BETA) where beta is the half angle, using the above result, and I'm totally stuck on that part.

6. Nov 23, 2006

### S_BX

Solved

Never mind guys - I got it eventually! :) Thanks for trying to help though! :)