The angle of elevation of the top of a tower height 100√3 m from a point 100 m from the foot of the tower on a horizontal plane is:

60° 30° 45° 75°

60°

AB = Tower = 100\(\sqrt {3}\) m BC = 100 m From figure, tanΘ = \(\frac{AB}{BC}\) = \(\frac{100\sqrt {3}}{100}\) = \(\sqrt {3}\) ⇒ Θ = 60°