Sina+sinc = 2 sin[(a+c)/2]* cos[(a-c)/2]= 2 sin[(π-b)/2]* cos(π/6)。

= (radical number 3)*sin(π/2-B/2)= radical number 3*cos(B/2)

sinB=2sin(B/2)*cos(B/2)。

The equation is (root number 3)*cos(B/2)=4sin(B/2)*cos(B/2).

Sin(B/2)= (root number 3)/4

Because a-c = π/3, a+c >; π/3，B& lt； 2π/3，B/2 & lt； π/3，cos(B/2)>0

So cos(B/2)== root number (1-sin (b/2) 2) = (root number 13)/4.

SinB=2sin(B/2)*cos(B/2)= (root number 39)/8.