The angle of elevation of the top of a tower of height 100 m from a point which is 100√3 m away from the foot of the tower on a horizontal plane is:
Answer & explanation
Correct answer: option 2
AB = Tower = 100 m
BC = 100\(\sqrt {3}\) m
From figure, tanΘ = \(\frac{AB}{BC}\) = \(\frac{100}{100\sqrt {3}}\) = \(\frac{1}{\sqrt {3}}\)
⇒ Θ = 30°