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#