Lets assume AB = 2√3 m, i.e. length of shadow on the ground from the pole

and BC = 6 m, i.e. height of the pole as mentioned in the diagram.

Let sun make an angle of θ on the ground, such that ∠CAB=θ

So in ∆ABC, tan θ = BC/AB

(∵ tan θ = perpendicular/base)

⇒ tan θ = tan 60° (∵ tan 60° = √3)

By comparing, we get

θ = 60°

∴, the sun’s elevation is 60°