If a pole 6 m high casts a shadow 2 m long on the ground then the sun’s elevation is

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 suns elevation is 60°