Solution: According to sine theorem A/SINA = B/SINB = C/SINC.
Get:
a =(Sina/sinB)* b；
c=(sinC/sinB)*b
Bring it into the known condition a+c = 2b.
Sina+Sina +sinC=2sinB available.
According to trigonometric functions and formulas
sinA+sinC = 2 sin[(A+C)/2]* cos[(A-C)/2]
∴A+B+C=∏
∫sin[(A+C)/2]= sin[(∏-B)/2]= sin(∏/2-B/2)= cos(B/2)
∴A-C=60
∫cos[(A-C)/2]= cos 30 =(√3)/2
∫sinA+sinC =√3 * cos(B/2)= 2 sinb
According to the angle doubling formula sinb = 2 sin (b/2) cos (b/2)
√3*cos(B/2)=4sin(B/2)cos(B/2)
sin(B/2)=(√3)/4；
cos(B/2)=√( 1-((√3)/4)^2)
=(√ 13)/4
Sinb = 2 sin (b/2) cos (b/2) = (√ 39)/8, and sinA is not needed.
I hope it will help you, hope to adopt it, thank you!