Question

已知在三角形ABC中,内角ABC所对的边分别是abc,且满足2asin(C+π/6)=b+c 求A的值

Answer 1

2asin(C+π/6) = b+c
根据正弦定理有：
2sinAsin(C+π/6) = sinB+sinC
2sinA{sinCcosπ/6+cosCsinπ/6) = sinB+sinC
sinA{√3sinC+cosC) = sinB+sinC

√3sinAsinC+sinAcosC = sinB+sinC

又，sinB=sin(A+C) = sinAcosC+cosAsinC
∴ √3sinAsinC+sinAcosC = sinAcosC+cosAsinC +sinC
∴ √3sinAsinC =cosAsinC +sinC
两边同除以sinC得：
√3sinA =cosA + 1

√3sinA - cosA = 1

2(sinAcosπ/6 - cosAsinπ/6) = 1

sin（A-π/6) = 1/2
A-π/6 ≠ 5π/6
∴ A-π/6 = π/6
∴ A = π/3