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High school mathematics trigonometric function examination paper
From vector m⊥ vector n, we get (√3c-2b)cosA+√3acosC=0, (changed the question).

∴√3(ccosA+acosC)=2bcosA,

∴√3b=2bcosA,

∴cosA=√3/2,A=π/6.

2.B=π/6=A,

∴b=a,C=2π/3,

According to the cosine theorem, the median line of BC is AM = √ [A2+(A/2) 2+A2/2] = (√ 7/2) A = √ 7,

∴a=2,

∴S△ABC=( 1/2)a^2*sinC=√3.