Recent content by S_BX

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

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...

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

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)...