Question

1.已知三角形ABC的面积S满足 根号3≤S≤3,且向量AB乘以向量BC等于6,向量AB与向量BC的夹角为θ.

Answer 1

面积S=(1/2)|AB||BC|sinθ,θ∈(0,π)

|AB||BC|=2S/sinθ

AB·BC=6

|AB||BC|cosθ=6>0,所以θ∈(0,π/2)

2Scosθ/sinθ=6

sinθ/cosθ=S/3

tanθ=S/3

√3 ≤S≤3

√3/3 ≤S/6≤1

√3/3 ≤tanθ≤1

θ∈(0,π/2)时,tanθ单调递增

故30°≤θ≤45°