在三角形中内角ABC的对边分别为abc?
的有关信息介绍如下:2bcosC=2a+c由正弦定理:a/sinA=b/sinB=c/sinC则2sinBcosC=2sinA+sinCA+B+C=π则A=π-(B+C)带入:2sinBcosC=2sin(B+C)+sinC=2sinBcosC+2cosBsinC+sinC则: 2cosBsinC+sinC=0因为C≠0,则sinC≠0, 则cosB=-1/2,则B=120°~~~~~~~~~~~~~~~~~~延长D至E,使得DE=BD,连接AE因为BD=DE,AD=CD,∠BDC=∠ADE则,可知△ADE≌△CDB则 AE=BC则∠BAE=∠B=120°在△ABE中,由余弦定理:c^2+a^2-2accosB=BE^2即:c^2+4-2*2*c*(-1/2)=(2√3)^2c=4 , c=-2(舍去)