Question

在三角形ABC中角A,B,C的对边分别为a,b,c,已知2B=A+C,a+根号二b=2c，求sinC的值

Answer 1

解答：
2B=A+C
3B=A+B+C=180°
∴ B=60°
a+√2b=2c
利用正弦定理
a/sinA=b/sinB=c/sinC
∴ sinA+√2sinB=2sinC
sinA=sin(B+C)=sin(60°+C)
∴ sin(60°+C)+√2*√3/2=2sinC
sin60°cosC+cos60°sinC+√6/2=2sinC
(3/2)sinC-(√3/2)cosC=√6/2
√3sin(C-π/6)=√6/2
∴ sin(C-π/6)=√2/2
∴ C-π/6=π/4
∴ C=5π/12
sinC=(√6+√2)/4