Question

已知三角形ABC的三个内角A.B.C对应的边长分别为a.b.c向量，向量m=(sinB,1-cosB)与向量n=(2,0)夹角阿法的余

已知三角形ABC的三个内角A.B.C对应的边长分别为a.b.c向量，向量m=(sinB,1-cosB)与向量n=(2,0)夹角阿法的余弦值1/2，求角B的大小，若三角形ABC外接圆半径为1，求a+c的范围 帮帮

Answer 1

向量m=(sinB,1-cosB)，向量n=(2,0)，
m•n=2sinB,
|m|=√(sin²B+(1-cosB) ²)=√(2-2 cosB)= √[2(1- cosB)]= √[2•2sin²(B/2)]=2 sin(B/2).
|n|=2
所以Cosα=m•n/(|m||n|)=2sinB/[4 sin(B/2)]= 4 sin(B/2)cos(B/2) /[4 sin(B/2)]= cos(B/2).
由已知：Cosα=1/2，
∴cos(B/2) =1/2，B/2 =π/3. B=2π/3.

由正弦定理得a/sinA=b/sinB=c/sinC=2R=2.
所以(a +c )/(sinA +sinC)=2
a +c=2(sinA +sinC)
∵B=2π/3. A +C=π/3.
∴a +c=2(sinA +sin(π/3-A))=2(sinA +√3/2cosA-1/2sinA)
=2(1/2sinA +√3/2cosA)=2sin (A+π/3)
因为0<A<π/3, π/3<A+π/3<2π/3.
所以√3/2<sin (A+π/3)≤1
a +c==2sin (A+π/3)∈（√3，2].