Problem about triangle

(a)Let angle OBA = x, angle CBA = y. Therefore x + y = 60 (given) or x = 60 - y ...(1)
(b) Let OB = m. For triangle AOB, by sine rule, m/sin(60 - x) = 3/sinx because angle BOA = 120. Or m/sin(60- x) = 3/sin(60- y).....................(2)
(c) For triangle COB, by sine rule, m/sin(60-y) = 4/siny or m/sin(60-y) = 4/sin(60-x)............(3)
(d) (2) x (3), we get m^2/[sin(60-x)(sin(60-y)] = 12/[sin(60-y)sin(60-x)],
therefore, m^2 = 12, therefore, m = BO = sqrt12.