在三角形ABC中,角A,B,C所对的边分别为a,b,c设b²+c²-bc=a²,c/b=1/2+根号3 1.求角A 2.

1.a² =b²+c²-2bccosA=b²+c²-bc cosA = 1/2,0＜A＜180° A = 60°
2.c/b=1/2+√3,代入 b²+c²-bc=a² 得 (15/4) b² = a² 故a/b = √ 15/2
a/sinA = b / sinB 得 sinB = 1/√5 cosB = 2/√5,tanB = 1/2