Question 391000
I have a circle inside a regular pentagon. CE is from top of pentagon to the bottom.  I know that CE is 1.4375. I need to find DE which is the radius of the circle. CE is a straight line going all the way through middle of the circle.
<pre>
{{{drawing(300,300,-1.1,1.1,-1.1,1.1,

line(-cos(3pi/10),-sin(3pi/10),cos(3pi/10),-sin(3pi/10)),
line(cos(3pi/10),-sin(3pi/10),cos(pi/10),sin(pi/10)),
line(cos(pi/10),sin(pi/10),0,1), circle(0,0,.02), locate(0.05,.05,D),
line(0,1,-cos(pi/10),sin(pi/10)), locate(0,1.1,C),
line(-cos(pi/10),sin(pi/10),-cos(3pi/10),-sin(3pi/10)),
locate(0,-cos(pi/5)-.03,E), 
red(circle(0,0,cos(pi/5))), green(line(0,1,0,-cos(pi/5))) )}}}

Draw DA and DB:

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,

line(-cos(3pi/10),-sin(3pi/10),cos(3pi/10),-sin(3pi/10)),
line(cos(3pi/10),-sin(3pi/10),cos(pi/10),sin(pi/10)),
line(cos(pi/10),sin(pi/10),0,1), circle(0,0,.02), locate(0.05,.05,D),
line(0,1,-cos(pi/10),sin(pi/10)), locate(0,1.1,C),
line(-cos(pi/10),sin(pi/10),-cos(3pi/10),-sin(3pi/10)),
locate(0,-cos(pi/5)-.03,E),locate(-cos(3pi/10),-sin(3pi/10)-.03,B),
locate(-1.03,.35,A),

red(circle(0,0,cos(pi/5))), green(line(0,1,0,-cos(pi/5))),

green(line(-cos(3pi/10),-sin(3pi/10),0,0), line(0,0,-cos(pi/10),sin(pi/10)))
 
)}}}

Angle ADC = {{{"360°"/5}}} = {{{"72°"}}}
Angle BDA = {{{"360°"/5}}} = {{{"72°"}}}

Therefore angle BDC = {{{"72°"+"72°"="144°"}}}

Erase DA (to keep figure from being cluttered)

Draw the diagonal CB, and label angle BDC as having measure 144°

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,

line(-cos(3pi/10),-sin(3pi/10),cos(3pi/10),-sin(3pi/10)),
line(cos(3pi/10),-sin(3pi/10),cos(pi/10),sin(pi/10)),
line(cos(pi/10),sin(pi/10),0,1), circle(0,0,.02), locate(0.05,.05,D),
line(0,1,-cos(pi/10),sin(pi/10)), locate(0,1.1,C),
line(-cos(pi/10),sin(pi/10),-cos(3pi/10),-sin(3pi/10)),
locate(0,-cos(pi/5)-.03,E),locate(-cos(3pi/10),-sin(3pi/10)-.03,B),
locate(-1.03,.35,A), locate(-.23,.05,"144°"),

red(circle(0,0,cos(pi/5))), green(line(0,1,0,-cos(pi/5)),line(0,0,-cos(3pi/10),-sin(3pi/10))),

green(line(-cos(3pi/10),-sin(3pi/10),0,1)) )}}}

Since triangle CDB is isosceles and has vertex angle 144°,
its two congruent base angles are 36° each because 180°-144°= 36° and
each of the base angles is one-half of that.  So we label angle BCD
as 18°:

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,

line(-cos(3pi/10),-sin(3pi/10),cos(3pi/10),-sin(3pi/10)),
line(cos(3pi/10),-sin(3pi/10),cos(pi/10),sin(pi/10)),
line(cos(pi/10),sin(pi/10),0,1), circle(0,0,.02), locate(0.05,.05,D),
line(0,1,-cos(pi/10),sin(pi/10)), locate(0,1.1,C),
line(-cos(pi/10),sin(pi/10),-cos(3pi/10),-sin(3pi/10)),
locate(0,-cos(pi/5)-.03,E),locate(-cos(3pi/10),-sin(3pi/10)-.03,B),
locate(-1.03,.35,A), locate(-.23,.05,"144°"),

red(circle(0,0,cos(pi/5))), green(line(0,1,0,-cos(pi/5)),line(0,0,-cos(3pi/10),-sin(3pi/10))),


locate(-.15,.55,"18°"),


green(line(-cos(3pi/10),-sin(3pi/10),0,1)) )}}}


CEB is a right triangle.  Therefore 

{{{BE/CE}}}=tan(18°)

BE = CE*tan(18°) and since CE is given as 1.4375

BE = 1.4375*tan(18°) = 0.4670720633

Angle BDE is suplementary to angle CDB which 144°,
so angle BDE is 180°-144° = 36°.   Now I'll erase some more:

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,

line(-cos(3pi/10),-sin(3pi/10),cos(3pi/10),-sin(3pi/10)),
line(cos(3pi/10),-sin(3pi/10),cos(pi/10),sin(pi/10)),
line(cos(pi/10),sin(pi/10),0,1), circle(0,0,.02), locate(0.05,.05,D),
line(0,1,-cos(pi/10),sin(pi/10)), locate(0,1.1,C),
line(-cos(pi/10),sin(pi/10),-cos(3pi/10),-sin(3pi/10)),
locate(0,-cos(pi/5)-.03,E),locate(-cos(3pi/10),-sin(3pi/10)-.03,B),
locate(-1.03,.35,A), locate(-.23,.05,"144°"),

red(circle(0,0,cos(pi/5))), green(line(0,1,0,-cos(pi/5)),line(0,0,-cos(3pi/10),-sin(3pi/10))),


locate(-.18,-.23,"36°") )}}}

Now triangle DBE is a right triangle, so

{{{BE/DB}}} = sin(36°)

BE = DB*sin(36°)

DB = {{{BE/sin("36°")}}} = {{{0.4670720633/sin("36°")}}} = 0.7946304565.

Since DB is a radius of the circle, that's what was required.

Answer radius = 0.7946 rounded to nearest ten thousandth.

Edwin</pre>