锐角三角形ABC中,a,b,c为角ABC所对的边,且(b-2c)cosA=a-2acos^2(B/2)
问题描述:
锐角三角形ABC中,a,b,c为角ABC所对的边,且(b-2c)cosA=a-2acos^2(B/2)
求角A.若a=根号3,求b+c的取值范围
答
(b-2c)cosA=a-2acos^2(B/2)则(sinB-2sinC)cosA=sinA-sinA(1+cosB)则sinBcosA-2sinCcosA=sinA-sinA-sinAcosBsinBcosA+cosBsinA-2sinCcosA=0sin(B+A)=2sinCcosAsinC=2sinCcosAcosA=1/2A=60°a=...