In an equilateral triangle, all its internal angles are equal to 60 degrees.
Now, we know that DC=#3x-10# and AE=#2x+5#
In #triangle AEC#,
#tan60=(AE)/(EC)#
#sqrt3=(2x+5)/(EC)#
#EC=1/sqrt3(2x+5)#
In #triangle AEB#,
#tan60=(AE)/(EB)#
#sqrt3=(2x+5)/(EB)#
#EB=1/sqrt3(2x+5)#
Therefore, we know that #BC=BE+EC=1/sqrt3(2x+5)+1/sqrt3(2x+5)=2/sqrt3(2x+5)#
Now, in #triangle BDC#,
#sin60=(DC)/(BC)#
#sqrt3/2=(3x-10)/(BC)#
#BC=2/sqrt3(3x-10)#
Since #BC=2/sqrt3(3x-10)# as well as #BC=2/sqrt3(2x+5)#, then we can say that
#2/sqrt3(3x-10)=2/sqrt3(2x+5)#
#3x-10=2x+5#
#x=15#
To find the actual length of BC, we can sub #x=15# back into #BC=2/sqrt3(3x-10)#
#BC=2/sqrt3(3times15-10)#
#BC=2/sqrt3times35#
#BC=70/sqrt3#
Finally, to find length of BF,
we can use #triangle BFC#,
#sin60=(BF)/(BC)#
#sqrt3/2=(BF)/(70/sqrt3)#
#BF=sqrt3/2times70/sqrt3#
#BF=70/2#
#BF=35#
