题目内容
在△ABC中,内角A,B,C所对的边分别为a,b,c,cos B=
.
(1)求cos(A+C)的值;
(2)求sin
的值;
(3)若
·
=20,求△ABC的面积.

(1)求cos(A+C)的值;
(2)求sin

(3)若


(1)-
(2)
(3)



(1)在△ABC中,∵A+B+C=π,∴A+C=π-B.
∵cos B=
,∴cos(A+C)=cos(π-B)=-cos B=-
.
(2)在△ABC中,∵cos B=
,∴sin B=
=
,
∴sin(B+
)=sin Bcos
+cos Bsin
=
×
+
×
=
.
(3)∵
·
=20,即|
|·|
|cos B=20,
∴c·a·
=20,即ac=25.
∴△ABC的面积S△ABC=
acsin B=
×25×
=
∵cos B=


(2)在△ABC中,∵cos B=



∴sin(B+








(3)∵




∴c·a·

∴△ABC的面积S△ABC=





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