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:
Answer & explanation
Correct answer: option 1
AB = Tower = 100\(\sqrt {3}\) m
BC = 100 m
From figure, tanΘ = \(\frac{AB}{BC}\) = \(\frac{100\sqrt {3}}{100}\) = \(\sqrt {3}\)
⇒ Θ = 60°