问题描述：
设三角形ABC的内角A,B,C,所对的边长分别为a,b,c,向量m=(cosA,cosC),向量n=(根号3c-2b,根号3a）,且向量m垂直向量n
1）求角A的大小
2）若角B=派\6,BC边上的中线AM的长为根号7,求三角形ABC的面积
问题解答：
我来补答
(1)
m=(cosA,cosC),n=(√3c-2b,√3a)
m垂直向量n
=>m.n=0
(cosA,cosC).(√3c-2b,√3a)=0
(√3c-2b)cosA+ √3a(cosC)=0
(√3c-2b)(b^2+c^2-a^2)/(2bc)+ √3(a^2+b^2-c^2)/(2b) =0
-(b^2+c^2-a^2)/c +(√3/(2b))(2b^2) =0
(b^2+c^2-a^2) = √3bc
a^2=b^2+c^2 -√3bc
by cosine rule
-√3bc = -2bc cosA
cosA =√3/2
A = π/6
(2)
B=π/6,M is mid point of BC
|AM| = √7
A=B => a=b
by sine rule
c/sinC = a/sinA
c = √3a
consider 三角形ABM
|AM|^2 = c^2 +(a/2)^2 - (ac)cosB
7 = 3a^2+a^/4 - (3/2)a^2
7 = 7a^2/4
a^2 = 4
a=2
三角形ABC的面积
=(1/2)absinC
=(1/2)a^2 ( √3/2)
=(1/2)4(√3/2))
= √3
