Question

在三角形ABC中，若向量AB·向量BC/3=向量BC·向量CA/2=向量CA·向量AB/1,则cosA=?

Answer 1

原式即 BA•BC / 3 = CB•CA / 2 = AC•AB
设△ABC的面积为S，则S=1/2*bc*sinA.
又因AC•AB = bc*cosA.将bc=2S/ sinA代入得：
AC•AB =2S / tanA
∴原式即 2S / 3tanB = 2S / 2tanC = 2S / tanA
即 3tanB = 2tanC = tanA
tanA = -tan(B+C) = -tanB + tanC) / ( tanB•tanC - 1 )
设tanA = x 则 tanB = x / 3 , tanC = x / 2
∴ x = ( x/3 + x/2 ) / ( x/3 • x/2 - 1 )
x = √11 , -√11(舍去)
∴ tanA = √11
所以cosA=1/√12=√3/6.