# Simplify for X? radian equation?

• January 6th 2011, 08:48 AM
DaveLevy
Arc Segments - Trig? Finding ratio (via golden angle)
Attachment 20346
This one has thrown me for a loop.
Basically I have to simplify and then solve for x where y is 1 and c is (1+√5)/2
and arc A = 2Pi(1 - 1/c)
2(arc A) + 2(arc B) equal a full circle.
I'm lost, any help is appreciated. Thank you.
• January 6th 2011, 09:33 AM
snowtea

Note that if $R$ was the radius, $Y = 2R\sin(B/2)$ and $X = 2R\sin(A/2)$.

So, $X=Y\frac{\sin(A/2)}{\sin(B/2)}$.

I think you should be able to go from there.
• January 6th 2011, 10:23 AM
DaveLevy
Using that I got A=2.3999632297287 and X = .38880073356013 if Y=1 ... I know this is not correct because x>y ... could you maybe expound upon that further simplifying A and B into the equation so I can see what's supposed to happen? thank you, I really appreciate your help

So basically I'm looking at it like this:
X=Y(cos(A/2)/cos(B/2)), Y=1, B=(2Pi-2A)/2, A=2Pi(1-1/c), c=(1+Root of 5)/2

Am I right about B at this first step?
• January 6th 2011, 10:50 AM
snowtea
Oops, it should have been sin and not cos.
• January 6th 2011, 12:37 PM
DaveLevy
Okay... so in that case I got x ~ 2.57... is that correct? :)

Edit:
Could someone help show me the step by step on how to solve this properly if you have the time and kindness? thank you.
• January 6th 2011, 01:49 PM
snowtea
I'm not that great adding drawings to posts, so I'll try to describe it.

To get $Y = 2R\sin(B/2)$, in the diagram, the rectangle is divided into 4 triangles. Look at one of these triangle with side length Y.
Bisect the angle of the triangle from the center of the circle, and you have two right triangles.
• January 6th 2011, 04:08 PM
DaveLevy
Trying to solve... ????
Attachment 20354
http://i818.photobucket.com/albums/z...veAttempt1.jpg
This is driving me nutty... okay so in that scenario the hypotenuse is the radius... however when I try to solve for X I get all wrong values... no matter which trig function I use... so what could I be doing incorrectly? :confused: