Question

三角形ABC中,边abc成等差数列,CosB=3/4.1.求(1/tanA)+(1/tanC)的值.2.设向量BA*BC

Answer 1

1.a+c=2b,

cos[(A-C)/2]=2sin(B/2),

1/tanA+1/tanC

=2sinB/sin(B/2)

=4cos(B/2)

=√14.

2.向量BA*BC=3ac/4=3/2,ac=2,

由余弦定理,

b^2=a^2+c^2-3ac/2=(a+c)^2-7,

把b=(a+c)/2代入上式,得

a+c=(2/3)√21.