Question

△ABC中,角A,B,C所对的边分别是abc,（2）若sinC/2sinA-sinC=b^2 -a^2 - c^2/c^

Answer 1

（1）∵在△ABC中,b2=a2+c2-2accosB,

∴b2-a2-c2=-2accosB,同理可得c2-a2-b2=-2abcosC

∵sinC/(2sinA−sinC)＝(b2−a2−c2)/(c2−a2−b2)

∴sinC/(2sinA−sinC)＝(−2accosB)/(−2abcosC)＝ccosB/bcosC＝sinCcosB/sinBcosC ,

∵sinC≠0,可得sinBcosC=2sinAcosB-sinCcosB,

∴2sinAcosB=sinBcosC+sinCcosB=sin（B+C）=sinA,

∵sinA≠0,∴等式两边约去sinA,可得cosB＝1/2 ,

∵0＜B＜π,∴角B的大小π/3