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
=4,则△ABC的面积是( )c cosC
A.
3
B. 4
C. 2
3
D. 2
答
∵
=a cosA
=b cosB
①,且由正弦定理得:c cosC
=a sinA
=b sinB
②,c sinC
∴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
=4,c cosC
∴a=b=c=2,即△ABC为边长是2的等边三角形,
则△ABC的面积S=
×2×2×sin60°=1 2
.
3
故选A