B/sinB= root number 3/2 root number 3=2
a = 2RsinA = 2sinA，c=2sinC
a b+ 2BC = C+2a = 2 sinc+4 Sina = 2 sinc+4 sin( 120-C)
=4sinC+2 root 3cosC, 4 square +(2 root 3) square =28, root 28=2 root 7.
Extract two root numbers 7: = (two root numbers 7)[(2/ root number 7)sinC+ (root number 3/ 7)cosC]
=(2 roots 7)sin(C+X), where sinX= root 3/ root 7 cosX=2/ root 7.
If the value range of sin(C+X) is-1 to 1, then (2 radicals 7)sin(C+X) is less than or equal to 2 radicals 7.
So the answer is the root number 7 of 2.
Supplement: asinC+bcosC=[ under the root number (Party A+Party B)] {[under the root number (Party A+Party B) ]sinC+[ under the root number (Party A+Party B) ]cosC}=[ under the root number (Party A+Party B)] SIN (C+X.
CosX=a/ radical sign (Party A+Party B)