Question

求△ABC的面积

△ABC的内角A,B C所对的边分别为a,b,c.向量m＝(a，√3 b)与n=(cosA,sinB)平行
1.求A 2.若a=√7,b=2,求△ABC的面积

Answer 1

(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