將a=2RsinA,b=2RsinB代入a2=b2+c2-2bccosA,得sin2A=sin2B+sin2C-2sinBsinCcosA,即sin2B+sin2C=sin2A+2sinBsinCcosA�、伲�?yàn)閏os2A+cos2B+cos2C=1,所以sin2A+sin2B+sin2C=2�、冢畬ⅱ俅擘�,得sin2A+(sin2A+2sinB·sinCcosA)=2,即-2cos2A+2sinBsinCcosA=0,所以2cosA(sinBsinC-cosA)=0,所以2cosA[sinBsinC+cos(B+C)]=0,所以cosAcosBcosC=0,所以cosA=0或cosB=0或cosC=0,所以△ABC是直角三角形.