第1个回答 2010-09-07
由条件把左边全移到右边得:
(sinA)^2+(sinC)^2+√3sinAsinC-(sinB)^2
=(sinA)^2+(sinC)^2+√3sinAsinC-[sin(A+C)]^2
=(sinA+sinC)^2+(√3-2)sinAsinC-[sin(A+C)]^2
=4[sin(A+C)/2]^2*[cos(A-C)/2]^2+(1-√3/2)[cos(A+C)-cos(A-C)]-[sin(A+C)]^2
=[cos(A+C)]^2-cos(A+C)cos(A-C)+√3cos(A-C)/2-√3cos(A+C)/2
=[cos(A+C)-cos(A-C)][cos(A+C)-√3/2]
=-2sinAsinC[cos(A+C)-√3/2]
=2sinAsinC(cosB+√3/2)
=0
即原式简化为sinAsinC(cosB+√3/2)=0,A、B、C为三角形三个角。
显然sinA和sinC不等于零,所以只能cosB+√3/2=0,求得B=150°