在锐角三角形ABC中,已知角A,B,C成等差数列,b=根号3,如果向量AB*向量BC=-3／2求1.三角形ABC的面积2.a+c的值

Question

在锐角三角形ABC中,已知角A,B,C成等差数列,b=根号3,如果向量AB*向量BC=-3／2
求1.三角形ABC的面积
2.a+c的值

Answer 1

let
B= A+d
C=A+2d
A+B+C =π
=> A+d =π/3
(1)
三角形ABC的面积
= (1/2)|AB||BC|sinB
= (1/2) (BA.BC) tanB
=(1/2) (3/2) tan(π/3) ( BA.BC = -AB.BC, B=A+d=π/3)
= (1/2) (3/2) √3
=3√3/4
(2)
AB.BC = -3/2
-ac ( cosB) =-3/2
ac = 3
By cosine-rule
b^2 = a^2 + c^2 -2acosB
= a^2+c^2 - ac
(a+c)^2 = b^2 + 3ac
= 3+9=12
a+c = 2√3