问题
解答题
在△ABC中,角A、B、C的对边分别为a、b、c,已知B=60°. (Ⅰ)若cos(B+C)=-
(Ⅱ)若a=5,
|
答案
(本小题满分12分)
(Ⅰ)在△ABC中,由cos(B+C)=-
,得11 14
sin(B+C)=
=1-cos2(B+C)
=1-(-
)211 14
,5 3 14
∴cosC=cos[(B+C)-B]=cos(B+C)cosB+sin(B+C)sinB
=-
×11 14
+1 2
×5 3 14
=3 2
;…(6分)1 7
(Ⅱ)由
•AC
=5,得|CB
|•|AC
|cos(180°-C)=5,即abcosC=-5,CB
又a=5,∴bcosC=-1,①
由正弦定理
=a sinA
,得b sinB
=a sin(120°-C)
,b sin60°
∴
=5
cosC+3 2
sinC1 2
,b 3 2
即
bcosC+bsinC=53
,②3
将①代入②,得bsinC=6
,3
则△ABC的面积为S=
absinC=1 2
×5×61 2
=153
.…(12分)3