We can solve this using circle theorem:

We know that XA=XB=XC=5m therefore the three sides are all radii of a circle with a radius of 5m
Therefore, we know:
2/_BCA=/_BXA
2/_ABC=/_AXC
2/_BAC=/_BXC
/_BXC=2(80)=160
/_AXC=2(60)=120
/_BXA=2(40)=80
Using the cosine we know that:
c^2=a^2+b^2-2bacosC
c=sqrt(a^2+b^2-2bacosC)
AB=sqrt(AX^2+XB^2-2(AX)(XB)cos(/_AXB))
color(white)(AB)=sqrt(5^2+5^2-2(5^2)cos(80))
color(white)(AB)~~6.43m
BC=sqrt(BX^2+XC^2-2(BX)(XC)cos(/_BXC))
color(white)(BC)=sqrt(5^2+5^2-2(5^2)cos(160))
color(white)(BC)~~9.89m
AC=sqrt(AX^2+XC^2-2(AX)(XC)cos(/_AXC))
color(white)(AC)=sqrt(5^2+5^2-2(5^2)cos(120))
color(white)(AC)~~8.66m
Sides:
AB~~6.43m
BC~~9.89m
AC~~8.66m