Question 1145898
<pre>

Alan has no clue as to how to solve an advanced trig problem like this,
so he tries to say yours is a foolish question.

Draw a right triangle ABC with angle A = 18°, opposite side = BC = x,
and hypotenuse = 1.  That makes sin(18°)=opposite/hypotenuse = x/1 = x.
So we want to find x.

{{{drawing(400,4000/13,-.1,1.2,-.5,.5,

locate(-.034,0.02,A),
locate(.97,0.02,B),
locate(.96,0.2,x=sin(18^o)),
locate(.97,0.33,C),
locate(.15,.08,18^o),
locate(.5,.23,1),
line(cos(pi/10),sin(pi/10),cos(pi/10),0),

line(0,0,cos(pi/10),sin(pi/10)),

line(0,0,cos(pi/10),0)


 )}}}

The angle C is 90°-18° = 72°, so we indicate that:

{{{drawing(400,4000/13,-.1,1.2,-.5,.5,

locate(-.034,0.02,A),
locate(.97,0.02,B),
locate(.96,0.2,x=sin(18^o)),
locate(.97,0.33,C),
locate(.15,.08,18^o),
locate(.5,.23,1),
line(cos(pi/10),sin(pi/10),cos(pi/10),0),

line(0,0,cos(pi/10),sin(pi/10)),

line(0,0,cos(pi/10),0),
locate(.85,.3,72^o)


 )}}}

We draw an identical right triangle ABD underneath:

{{{drawing(400,4000/13,-.1,1.2,-.5,.5,

locate(.96,-0.1,x=sin(18^o)),

locate(-.034,0.02,A),
locate(.97,0.02,B),
locate(.96,0.2,x=sin(18^o)),
locate(.97,0.33,C),
locate(.15,.08,18^o),
locate(.5,.23,1),

locate(.85,.3,72^o), locate(.85,-.2,72^o),

locate(.15,.019,18^o),


line(cos(pi/10),sin(pi/10),cos(pi/10),0),

line(0,0,cos(pi/10),sin(pi/10)),

locate(.97,-0.3,D),

line(0,0,cos(pi/10),0),

line(cos(pi/10),sin(pi/10),cos(pi/10),0),

line(cos(pi/10),-sin(pi/10),0,0),
line(cos(pi/10),-sin(pi/10),cos(pi/10),0)



 )}}}

Next we draw CE to bisect the 72° angle C into
two 36° angles.

Let's let DE = y

{{{drawing(400,4000/13,-.1,1.2,-.5,.5,

locate(-.034,0.02,A),
locate(.97,0.02,B),
locate(.96,0.2,x=sin(18^o)),
locate(.97,0.33,C),
locate(.15,.08,18^o),
locate(.5,.23,1),

locate(.97,-0.3,D),
locate(.57,-.19,E),
locate(.15,.019,18^o),
locate(.77,-.245,y),
locate(.26,-.1,1-y),

locate(.96,-0.1,x=sin(18^o)),
locate(.8,.28,36^o),
locate(.87,.23,36^o),

locate(.85,-.2,72^o), 
line(0,0,cos(pi/10),sin(pi/10)),
line(0,0,cos(pi/10),-sin(pi/10)),
line(0,0,cos(pi/10),0),
line(cos(pi/10),sin(pi/10),cos(pi/10),-sin(pi/10)),
line(cos(pi/10),sin(pi/10),.5877852525,-.1909830057)



 )}}}

Now triangles CDE is similar to triangle ACD, because they
each have a 36° angle and a 72° angle.  Also they are both isosceles.
Triangle ACE is also isosceles because it has a 36° angle at both A 
and C.  So AE=CE=CD.

Since AC=1=AD, AE = 1-y.  By similar triangles,

{{{matrix(2,3,

AC/CD,""="",CD/DE,
1/(2x),""="", (2x)/y))}}}

Cross-multiply:

{{{matrix(6,3,
y,""="",4x^2,
AE,""="",CD,
1-y,""="",2x,
1-4x^2,""="",2x,
-4x^2-2x+1,""="",0,
4x^2+2x-1,""="",0))}}}

Use the quadratic formula:

{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}}
{{{x = (-2 +- sqrt(2^2-4(4)(-1) ))/(2(4))}}}
{{{x = (-2 +- sqrt(4+16))/8}}}
{{{x = (-2 +- sqrt(20))/8}}}
{{{x = (-2 +- sqrt(4*5))/8}}}
{{{x = (-2 +- 2sqrt(5))/8}}}
{{{x = (2(-1 +- sqrt(5)) )/8}}}
{{{x = (-1 +- sqrt(5))/4}}}

We take the + since the - sign gives a negative answer.

{{{x = (-1 + sqrt(5))/4}}}

{{{sin(18^o) = ( sqrt(5)-1)/4 }}}

Edwin</pre>