Question

已知a ，b，c分别是三角形ABC三个内角A，B，C的对边，且2asin（C＋π╱3）＝√3乘上b

Answer 1

1.
由正弦定理得2sinAsin(C+π/3)=√3sinB
2sinA(sinCcosπ/3+cosCsinπ/3)=√3sin(A+C)
sinAsinC+√3sinAcosC=√3sinAcosC+√3cosAsinC
sinC(sinA-√3cosA)=0
sinC恒>0，因此sinA-√3cosA=0
(1/2)sinA-(√3/2)cosA=0
sin(A-π/3)=0
A=π/3
2.
a=BC，b=AC，c=AB=3，d=BD=√13，令AD=x
由余弦定理得cosA=(c²+x²-d²)/(2cx)
A=π/3，c=3，d=√13代入，整理，得x²-3x-4=0
(x+1)(x-4)=0
x=-1(舍去)或x=4
b=2x=8
S△ABC=½bcsinA
=½·4·3·sin(π/3)
=½·4·3·(√3/2)
=3√3

Answer 2

解：
1.
2asin(c+π/3)=√3b
由正弦定理得2sinasin(c+π/3)=√3sinb
2sina[sinccos(π/3)+coscsin(π/3)]=√3sin(a+c)
2sina[(1/2)sinc+(√3/2)cosc]=√3(sinacosc+cosasinc)
sinasinc+√3sinacosc=√3sinacosc+√3cosasinc
sinc(sina-√3cosa)=0
2sinc[(1/2)sina-(√3/2)cosa]=0
2sincsin(a-π/3)=0
c为三角形内角，sinc>0，因此只有sin(a-π/3)=0
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