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Solve the combination of triangle and vector It is known that a, B and C are the opposite sides of three internal angles a, B and C of Δ ABC, vector m (radical 3, - 1) and vector n (COSA, Sina). If vector m, vector N and acosb + bcosa = csinc, then angle B =?)
In a triangle, so all angles belong to (0, π)
acosB+bcosA=csinC
According to the sine theorem, a = 2R · Sina, B = 2R · SINB, C = 2R · sinc can be substituted
2R·sinA·cosB+2R·sinB·cosA=2R·sinC·sinC
sinA·cosB+sinB·cosA=sin²C
sin（A+B）=sin²C
sin（180-C）=sin²C
sinC=sin²C
The solution is sinc = 0 (rounding) or sinc = 1, so C = π / 2
∵ m, n
∴m·n=0
That is √ 3cosa Sina = 0
2sin[A-（π/3）]=0
A=π/3
In conclusion, B = π / 6
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