Trigonometry: Solving for Cosine Using Segment Angle and Triangle Properties

  • Thread starter Thread starter S_BX
  • Start date Start date
  • Tags Tags
    Angle Trig
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
5 replies · 4K views
S_BX
Messages
4
Reaction score
0
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:

Discussion

Physics news on Phys.org
You could use the Pythagorean Theorem to find the 3rd side. But, there is too much information, so [tex]\cos\beta[/tex] can be expressed many ways.
 
lotrgreengrapes7926 said:
You could use the Pythagorean Theorem to find the 3rd side. But, there is too much information, so [tex]\cos\beta[/tex] can be expressed many ways.
I just took the 3rd side to be R-h.
 
Integral said:
Could you show me how you came up with

[tex]\frac {L^2 + 4h^2} {8h}[/tex] 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.
 
Solved

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