数学急!在ABC中,ABC对应a,b,c,向量m=(b,a-2c),

(1)
m⊥n
m.n=0
(b,a-2c).(cosA-2cosC,cosB)=0
b(cosA-2cosC)+(a-2c)cosB=0
b[(b^2+c^2-a^2)/(2bc)- 2(a^2+b^2-c^2)/(2ab)] +(a-2c)[(a^2+c^2-b^2)/(2ac)]=0
b[a(b^2+c^2-a^2)-2c(a^2+b^2-c^2)] +b(a-2c)(a^2+c^2-b^2) =0
ab(2c^2)-2bc(2a^2)=0
c-2a=0
c=2a
by sine rule
a/sinA = c/sinC
sinA/sinC = a/c = a/(2a) = 1/2
(2)
a=2
=> c= 4
|m|=3√5
=> b^2+(a-2c)^2=45
b^2+a^2+4c^2-4ac =45
b^2 +4+ 64-32=45
b= 3
cosA = (b^2+c^2-a^2)/(2bc)
= (9+16-4)/(24)
= 7/8
sinA = √15/8
Area
=(1/2)bcsinA
=(1/2)(3)(4) √15/8
=3√15/4