已知向量m=(sinA,sinB),n=(cosB,cosA),m•n=sin2C,且A、B、C分别为△ABC的三边a、b、c所对的角,(Ⅰ)求角C的大小;(Ⅱ)若sinA,sinB,sinC成等差数列,且CA•(AB-AC)=18,求c边

问题描述:

已知向量

m
=(sinA,sinB),
n
=(cosB,cosA),
m
n
=sin2C,且A、B、C分别为△ABC的三边a、b、c所对的角,
(Ⅰ)求角C的大小;
(Ⅱ)若sinA,sinB,sinC成等差数列,且
CA
•(
AB
-
AC
)=18,求c边的长及△ABC的面积.

(1)

m
n
=sinAcosB+cosAsinB=sin(A+B)=sin2C,
∴sinC=sin2C=2sinCcosC,
∴cosC=
1
2

∵C∈(0,π),∴C=
π
3

(2)∵sinA,sinB,sinC成等差数列,
∴sinA+sinC=2sinB,
由正弦定理可知a+b=2c,
又∵
CA
•(
AB
-
AC
)=18,
CA
CB
=18
,∴abcos
π
3
=18
,即ab=36.
由余弦定理得:c2=a2+b2-2abcosC=(a+b)2-3ab=4c2-108,
∴c2=36,解得c=6.
S△ABC=
1
2
absinC
=
1
2
×36×
3
2
=9
3