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.
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.
[Edit]
Note that if $\displaystyle R$ was the radius, $\displaystyle Y = 2R\sin(B/2)$ and $\displaystyle X = 2R\sin(A/2)$.
So, $\displaystyle X=Y\frac{\sin(A/2)}{\sin(B/2)}$.
I think you should be able to go from there.
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?
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.
I'm not that great adding drawings to posts, so I'll try to describe it.
To get $\displaystyle Y = 2R\sin(B/2)$, 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.