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°

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