在△ABC中,若2sinB=sinA+sinC,且A-C=π/2,则a:b:c=

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在△ABC中,若2sinB=sinA+sinC,且A-C=π/2,则a:b:c=在△ABC中,若2sinB=sinA+sinC,且A-C=π/2,则a:b:c=在△ABC中,若2sinB=sinA+s

在△ABC中,若2sinB=sinA+sinC,且A-C=π/2,则a:b:c=
在△ABC中,若2sinB=sinA+sinC,且A-C=π/2,则a:b:c=

在△ABC中,若2sinB=sinA+sinC,且A-C=π/2,则a:b:c=
A-C=π/2,故sinA=sin(π/2+C)=cosC+sinC*0=cosC,cosA=-sinC(A为钝角,C为锐角).
又B=π-(A+C),有sinB=sin(A+C)=sinAcosC+cosAsinC=(cosC)^2-(sinC)^2=cos(2C)
因此,可以得出:2(cosC)^2-2(sinC)^2=sinA+sinC=cosC+sinC
又cosC+sinC不为0,故:cosC-sinC=1/2
根据(sinC)^2+(cosC)^2=1可得cosC*sinC=3/8,由此可得出cosC+sinC=√7/4
求解可得出:sinC=(√7/4-1/2)/2=(√7-2)/8,sinA=cosC=(√7+2)/8,sinB=(sinA+sinC)/2=√7/8
故:a:b:c=sinA:sinB:sinC=(√7+2)/8:(√7-2)/8:√7/8=(√7+2):(√7-2):√7
给个精彩回答吧.