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设三角形ABC的内角A.B.C的对边分别为abc且c=2b,向量M=(sinA,3/2),N=(1,sinA+根3cos

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设三角形ABC的内角A.B.C的对边分别为abc且c=2b,向量M=(sinA,3/2),N=(1,sinA+根3cosA)
m n共线 求a/c 的值
设三角形ABC的内角A.B.C的对边分别为abc且c=2b,向量M=(sinA,3/2),N=(1,sinA+根3cos
m n共线
=> sinA/ (3/2) = 1/(sinA+√3cosA)
sinA((sinA+√3cosA)) = 3/2
(sinA)^2 + (√3/2)sin2A = 3/2
(1-cos2A)/2+ (√3/2)sin2A = 3/2
(√3/2)sin2A -(1/2)cos2A = 1
sin(2A-π/6)=1
2A-π/6 = π/2
A = π/3
c=2b
c/sinC = b/sinB
sinB/sinC = b/c = 1/2
A+B+C= π
B = 2π/3 -C
sinB = (√3/2)cosC + (1/2)sinC
sinB/sinC = √3/(2 tanC)+1/2
0= √3/(2 tanC)
cosC = 0
C = π/2
by sine-rule
a/sinA =c/sinC
a/c = sinA/sinC = √3/2