The measure of angle of elevation of top of tower 75√3 m high from a point at a distance of 75m from foot of tower in a horizontal plane is: Share with your friends Share 1 Varun.Rawat answered this Let AB be the tower. Let C be the point of observation.Let ∠ACB = θNow, AB = 753 m; BC = 75 mtan θ = perpendicularbase = ABBC⇒tan θ = 75375⇒tan θ = 3⇒θ = 60° 0 View Full Answer