Trigonometry: Solving for Cosine Using Segment Angle and Triangle Properties

[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

 

[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.

 

[S_BX]

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.

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

 

[Integral]

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?

 

[S_BX]

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?

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.

 

[S_BX]

Solved

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