在△ABC中,a,b,c分别是角A,B,C对边的长,且满足cosB?bcosC+2a+c=-b2a+c.(1)求角B的值.(2)若b=

2025-04-04 05:59:56
推荐回答(1个)
回答1:

(1)由已知得:2acosB+ccosB-2ab-bc=-bcosC-2ab-bc,
∴2acosB+ccosB+bcosC=0,
由正弦定理得,2sinAcosB+sinCcosB+cosCsinB=0,
整理得:2sinAcosB+sin(B+C)=0,即2sinAcosB+sinA=0,
∵sinA≠0,
∴cosB=-

1
2

又0<B<180°,
∴B=120°;
(2)在△ABC中,b=7,a+c=8,cosB=-
1
2

∴由余弦定理得,72=a2+c2+ac,
∴49=(a+c)2-ac,即49=82-ac,
∴ac=15,
∴S△ABC=
1
2
acsinB=
1
2
×15×sin120°=
15
3
4

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