已知三角形ABC的三个内角A.B.C所对的边分别是a.b.c，向量m=(1,1
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已知三角形ABC的三个内角A.B.C所对的边分别是a.b.c，向量m=(1,1-√3sinA),n=(cosA,1)，且m⊥n.(1)求角A；(2)若b+c=√3a，求sin(B+π/6)的值。
解：(1)m·n = cosA +1 - √3 sinA
= 1 - 2 sin(A - π/6)
= 0
sin(A- π/6) = 1/2
-π/6 < A - π/6 < 5π/6
∴A - π/6 = π/6 A = π/3
(2) b+c = √3 a
由正弦定理 sinB + sinC = √3 sinA = 3/2
sinC = sin(2π/3 -B) = √3 /2 cosB + 1/2 sinB
∴ 1/2 cosB + √3/2 sinB = √3/2
sin(B+π/6) = √3 /2
解：(1)m·n = cosA +1 - √3 sinA
= 1 - 2 sin(A - π/6)
= 0
sin(A- π/6) = 1/2
-π/6 < A - π/6 < 5π/6
∴A - π/6 = π/6 A = π/3
(2) b+c = √3 a
由正弦定理 sinB + sinC = √3 sinA = 3/2
sinC = sin(2π/3 -B) = √3 /2 cosB + 1/2 sinB
∴ 1/2 cosB + √3/2 sinB = √3/2
sin(B+π/6) = √3 /2
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