设△ABC的内角A,B,C所对边分别为a,b,c,且a+c=6,b=2, cosB= 7 9 .(1)求a,c的值;(2

2025-06-27 21:35:32
推荐回答(1个)
回答1:

(1)∵a+c=6①,b=2,cosB=
7
9

∴由余弦定理得:b 2 =a 2 +c 2 -2accosB=(a+c) 2 -2ac-
14
9
ac=36-
32
9
ac=4,
整理得:ac=9②,
联立①②解得:a=c=3;
(2)∵cosB=
7
9
,B为三角形的内角,
∴sinB=
1-(
7
9
)
2
=
4
2
9

∵b=2,a=3,sinB=
4
2
9

∴由正弦定理得:sinA=
asinB
b
=
4
2
9
2
=
2
2
3

∵a=c,即A=C,∴A为锐角,
∴cosA=
1-si n 2 A
=
1
3

则sin(A-B)=sinAcosB-cosAsinB=
2
2
3
×
7
9
-
1
3
×
4
2
9
=
10
2
27