Question

这道题中向量AB-向量BC不就是等于向量CA吗就是等于2呀为什么要选D呢

Answer 1

图

追问

是第8题

Answer 2

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.