Question

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.

2.求tanB

Answer 1

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

Answer 2

1. b²+c²-bc=a² => cosA = 1/2, A = 60°
2. c/b=1/2+√3, 代入 b²+c²-bc=a² =》(15/4) b² = a² => a/b = √ 15/2
正弦定理： a/sinA = b / sinB => sinB = 1/√5
=> cosB = 2/√5, tanB = 1/2

Answer 3

a² =b²+c²-2bccosA=b²+c²-bc
cosA = 1/2
所以 A = 60°
c/b=1/2+√3 因为b²+c²-bc=a²
得 (15/4) b² = a² 所以a/b = √ 15/2
a/sinA = b / sinB
得 sinB = 1/√5
cosB = 2/√5, tanB = 1/2