Question

一道数学题，急求解答！

Answer 1

SinC/c
=
SinA/a;
2acos(π/3+C)
=
a*CosC-2a*Sin(π/3)*SinC=a*CosC-2c*Sin(π/3)*SinA;
余弦定理得
CosC=(a^2+b^2-c^2)/2ab
因此
b-c=(a^2+b^2-c^2)/2b-2c*Sin(π/3)*SinA;
(b^2+c^2-a^2)/2bc
+
2*Sin(π/3)*SinA=1;
CosA+2*Sin(π/3)*SinA=1;
Cos(π/3)*CosA+Sin(π/3)*SinA=0.5;
因为在三角形中,所以
Cos(A-π/3)=0.5;
A
-
π/3=
π/3;
A=2*π/3;