www.fltk.net > 设△ABC的内角A，B，C 所对边分别为A,B ,C 且BCosC=A%2/1C 若B=1 △AB

 设△ABC的内角A，B，C 所对边分别为A,B ,C 且BCosC=A%2/1C 若B=1 △AB 发布于:2020-04-08 06:41:16

（1）在△ABC中，有sinA=sin（B+C）=sinBcosC+cosBsinC，由正弦定理得：

（1） （2） (1)∵a=bcosC+csinB∴由正弦定理知sinA=si

（1）∵a=bcosC+csinB，∴根据正弦定理，得sinA=sinBcosC+sinBsinC…

（1） 利用正弦定理:a/sinA=b/sinB=c/sinC ∵ a=bcosC+csinB ∴

(1)求角B大小(2)若b=1.求三角形ABC周长l的取值范围 bcosC=a-1/2c bcosC

（1） 利用正弦定理:a/sinA=b/sinB=c/sinC ∵ a=bcosC+csinB

（1）bcosC=(2a-c)cosB b(a＾2+b＾2-c＾2)/2ab=(2a-c)(a＾
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