设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,且bcosC=

设三角形ABC外接圆半径为r
则2r=a/sinA =b/sinB=c/sinC
(1)b^2=a^2+c^2-2accosB
cosB=(a^2+c^2-b^2)/2ac
cosC=(a^2+b^2-c^2)/2ab
b*(a^2+b^2-c^2)/2ab =(2a-c)(a^2+c^2-b^2)/2ac
a^2+b^2-c^2=(2a-c)(a^2+c^2-b^2)/c=2a(a^2+c^2-b^2)/c -a^2-c^2+b^2
2a^2=2a(a^2+c^2-b^2)/c
a=(a^2+c^2-b^2)/c
(a^2+c^2-b^2)/2ac=cosB=1/2
得B=60
（2)
A+C=120
30<A<90 30<C<90
C=120-A
sin^2 C=sin^2 (120-A)=(sin120cosA-cos120sinA)^2
=(根号3/2 cosA+1/2sinA)^2
=3/4cos^2 A+1/4 sin^2 A+根号3/2 sinAcosA
所以
sin^2 A+sin^2 C
=5/4 sin^2 A+3/4 cos^2 A+根号3/2 sinAcosA
=5/4 *(1-cos2A)/2 +3/4 *(1+cos2A)/2 +根号3/4 sin2A
=5/8-5/8cos2A+3/8+3/8cos2A+根号3/4 sin2A
=1-1/4cos2A+根号3/4 sin2A
=1+1/2(sin(-30)cos2A+cos(-30)sin2A)
=1+1/2 sin(-30+2A)
30<2A-30<150
所以1<1+1/2sin(-30+2A)<=3/2
所以1<sin^2 A+sin^2 C<=3/2

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