a=b或4,由题知道 B=60或120
1.当B=60时C=120-A,由
2sinCcosA=sinB知道
2sin(120-A)cosA=sinB
强前面式子展开后可得
2[sin120cosA-cos120sinA]cosA=2sin120cos^2(A)-2cos120sinAcosA
=sin120[1+cos2A]-cos120sin2A=sinB=sin60
即sin120cos2A-cos120sin2A=sin(120-2A)=0
因cosA>0,so A
a=b或4,由题知道 B=60或120
1.当B=60时C=120-A,由
2sinCcosA=sinB知道
2sin(120-A)cosA=sinB
强前面式子展开后可得
2[sin120cosA-cos120sinA]cosA=2sin120cos^2(A)-2cos120sinAcosA
=sin120[1+cos2A]-cos120sin2A=sinB=sin60
即sin120cos2A-cos120sin2A=sin(120-2A)=0
因cosA>0,so A