CUET Preparation Today
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: |
60° 30° 45° 75° |
30° |
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° |
