Given the length of the waist midline of an isosceles triangle is 3 under the root sign, what is the maximum area of the triangle?

Analysis: Let the waist length x of triangle be θ.

∵ waist centerline √3.

∴ Through cosine theorem 3 = 5/4x2-X2COS θ = > X 2 =12/(5-4COS θ)

∴ Triangle area =1/2x2sinθ = 6s inθ/(5-4cosθ)

Let f(θ)=6sinθ/(5-4cosθ)

Let f' (θ) = (30cosθ-24)/(5-4cosθ) 2 = 0 = >; cosθ=4/5

When θ= arccos4/5, F(θ) takes the maximum value.

When the vertex angle of the triangle is arccos4/5, the maximum area is 2.