解：ED=sin(π/3)/sin(2π/3-θ)*BD=sin(π/3)/sin(2π/3-θ) FD=sin(π/3)/sin(π/6+θ)*AD=sin(π/3)/sin(π/6+θ) 得FD/ED=sin(2π/3-θ)/sin(π/6+θ)=√3/2 即sin(π/3+θ)/sin(π/6+θ)=√3/2 展开 可解得 tanθ=√3 θ=π/3 希望能帮到你！
