已知A、B、C为△ABC的三内角,且其对边分别为a、b、c,若cosBcosC-sinBsinC=12.(Ⅰ)求A; (Ⅱ)若a=2

2024-10-30 23:27:13
推荐回答(1个)
回答(1):

(Ⅰ)∵cosBcosC?sinBsinC=

1
2

cos(B+C)=
1
2

又∵0<B+C<π,∴B+C=
π
3

∵A+B+C=π,∴A=
3

(Ⅱ)由余弦定理a2=b2+c2-2bc?cosA
得 (2
3
)2=(b+c)2?2bc?2bc?cos
3

即:12=16?2bc?2bc?(?
1
2
)
,∴bc=4,
S△ABC
1
2
bc?sinA=
1
2
?4?
3
2
3