Question

急！求解一道高中数学题。（麻烦附上详细过程，谢谢TAT）

Answer 1

原式可化为：
2sin(A+π/3)=2sinB
sin(A+π/3)=sinB
a≥b==>A≥B
sin(A+π/3)=sin(π-B)
A+π/3=π-B
A+B=2π/3
C=π/3
(2)
由正弦定理：
(a+b)/c=[sinA+sinB]/sinC
=[sinA+sin(2π/3-A)]/(√3/2)
=[sinA+(√3/2cosA+1/2sinA)]/(√3/2)
=[(3/2)sinA+(√3/2)cosA]/(√3/2)
=[(√3)sinA+cosA]
=2sin(A+π/6)
0<A<2π/3
π/6<A+π/6<5π/6
1/2＜sin(A+π/6)≤1
1＜2sin(A+π/6)≤2
1<(a+b)/c≤2