Now, I want contiune my work.
In my last post ,we got cos 72 degrees =(-1+$${\sqrt{}}$$5)/4
Any interior angle of regular pentagon=(5-2)*180/5 degrees=108 degrees
cos 108 degrees =-cos($${\frac{{\mathtt{\pi}}}{{\mathtt{2}}}}$$-72 degrees) =-cos72 degrees=(1-$${\sqrt{}}$$5)/4
In any pentagon ,every sides are equal and every diagonals are equal.
Lets say we are fing the ratio between AB and AC
According to law of cosine
In triangle ABC,we have AB^2=AC^2+BC^2-2*AC*BC*cosC, rearrange we have
2*AC*BC*cosC=AC^2+BC^2-AB^2
and because AC and BC are the sides of regular pentagon,so
2*AC^2*cosC=2*AC^2-AB^2
Both side divide by 2*AC^2,then we get
cosC=1-AB^2/2*AC^2
AB^2/2*AC^2=1-cosC
AB^2/2*AC^2=1-(-cos72 degrees)
AB^2/2*AC^2=1+cos72 degrees
AB^2/2*AC^2=1+(-1+5)/4
$${\frac{\left({\mathtt{sqrt5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}$$
Now,I am going to rearrange the Alan's equation and check
sin18 degrees=s/2d
2*sin 18 degrees=s/d
1/(2*sin18 degrees)=d/s
d/s=1/(2*sin18 degrees)=1.6180339887496196