Question

已知△ABC中，角A,B,C所对的边分别是a,b,c且a=4,b+c=5,√3sinC-cosC=1，求△ABC的面积

Answer 1

√3sinC-cosC=1
2sin(C-π/6)=1
C-π/6=π/6
C=π/3

cosC=1/2=(a²+b²-c²)/2ab
a=4,c=5-b
所以16+b²-25+10b-b²=4b
6b=9
b=3/2
所以S=1/2absinC=3√3/2

Answer 2

√3sinC-cosC=2 * sin（C-π/6）=1
C-π/6=π/6，C=π/3
过A做AD⊥BC于D
设DC为x
那么有BD^2+AD^2=AB^2
即（4-x）^2+3x^2=(5-2x)^2
解得x=3/4
所以高h=√3x=3√3/4
面积s=(AD*BC)/2=3√3/2

Answer 3

S△ABC=3

Answer 4

√3sinC-cosC=1
√3/2sinC-1/2cosC=1/2
sinCcos(π/6)-cosCsin(π/6)=1/2
sin(C-π/6)=1/2
C-π/6=π/6
C=π/3
sinC=sin(π/3)=√3/2
cosC=(a²+b²-c²)/2ab=1/2
a²+b²-c²=ab
16+(b+c)(b-c)=4b
16+5(b-c)=4b
b-5c+16=0
b-5(5-b)+16=0
6b=9
b=3/2
三角形面积S=1/2absinC=1/2*4*3/2*√3/2=3√3/2

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