设a,b,c分别是△ABC的三个内角A,B,C所对的边,且满足acosA=bcosB=ccosC=4,则△ABC的面积是(  ) A.3 B.4 C.23 D.2

问题描述:

设a,b,c分别是△ABC的三个内角A,B,C所对的边,且满足

a
cosA
b
cosB
c
cosC
=4,则△ABC的面积是(  )
A.
3

B. 4
C. 2
3

D. 2

a
cosA
=
b
cosB
=
c
cosC
①,且由正弦定理得:
a
sinA
=
b
sinB
=
c
sinC
②,
∴①÷②得:tanA=tanB=tanC,又A,B,C都为三角形的内角,
∴A=B=C=60°,又
a
cosA
=
b
cosB
=
c
cosC
=4,
∴a=b=c=2,即△ABC为边长是2的等边三角形,
则△ABC的面积S=
1
2
×2×2×sin60°=
3

故选A