在△ABC中，A,B,C的对边为a,b,c.且a,b,c成等比数列，求：①∠B的范围②求f(B)=sinB+根号3cosB的最值

a/sinA=b/sinB=c/sinC
b^2=ac
(sinB)^2=sinA*sinC=-（cos(A+C)-cos(A-C)）/2=(cosB+cos(A-C))/2
令cosB=x
1-x^2=(x+cos(A-C))/2
2x^2+x+cos(A-C)-1=0
2(x+1/4)^2+cos(A-C)-9/8=0
2(x+1/4)^2=9/8-cos(A-C)
0<=cos(A-C)<=1
1/16<=(x+1/4)^2<=9/16
1/4<=|x+1/4|<=3/4
由于B<90°（b为第二长的边）
cosB>0
0<cosB<=1/2
故B取值范围为60°<=B<90°
2、
f(B)=sinB+3^(1/2)cosB
=2sin(B+60°）
60°<=B<90°
1<f(B)<=3^(1/2)