Draw a circle of a radius AB=10 with a center A and choose on it a point and label it B.
From point B, using the same radius 10, draw an arc that intersects a circle at point F.
Obviously, Delta ABF is equilateral triangle, AB=BF=AF=10 and /_BAF=60^o.
Bisect /_BAF by a radius AD, so /_BAD=/_DAF=/_30^o.
Bisect /_BAD by a radius AC, so /_BAC=/_CAD=/_15^o.
We are ready now to derive the length BC from BD, which, in turn, we can derive from BF, that we know is equal to 10.
Let ADnnBF=P
Using Pythagorean Theorem, calculate BD from BP and PD.
BD^2 = BP^2+PD^2
BP = (BF)/2 = 10/2 = 5
PD = AD - AP
AD = 10
AP^2 = AF^2 - PF^2 = 10^2-(10/2)^2 = 100-25=75
AP = sqrt(75) = 5sqrt(3)
PD = 10 - 5sqrt(3)
BD^2 = (10/2)^2+(10-5sqrt(3))^2=25+100-100sqrt(3)+75=
=100(2-sqrt(3))
BD = 10sqrt(2-sqrt(3)) ~~5.176381
Analogously, calculate BC by knowing BD.
Let ACnnBD=Q
Using Pythagorean Theorem, calculate BC from BQ and QC.
BQ = (BD)/2 = 5sqrt(2-sqrt(3))
QC = AC - AQ
AC=10
AQ^2=AB^2-BQ^2=10^2-25(2-sqrt(3))=25(2+sqrt(3))
AQ = 5sqrt(2+sqrt(3))
QC=10-5sqrt(2+sqrt(3))
BC^2=BQ^2+QC^2=
=25(2-sqrt(3))+100-100sqrt(2+sqrt(3))+25(2+sqrt(3))=
=100(2-sqrt(2+sqrt(3)))
BC=10sqrt(2-sqrt(2+sqrt(3)))~~2.610524
