Let the Distance of the object from the tower be x meters.

∴ BC = x m

Given –

height of tower = AB = 60 m

Angle of depression = ∠ DAC = 30°

∴ ∠ BCA = ∠ DAC = 30°

[∵ When two ‖ lines are intersected by a third line then the Alternate interior angles will be equal.]

Now, In Δ ABC

tan 30° = AB/BC = 60/x [∵ tan θ = perpendicular/base]

⇒ 1/√3 = 60/x

∴ x = 60√3 meters