Mohit is standing at some distance from a 60 meters tall building. Mohit is 1.8 meters tall. When Mohit walks towards the building, then the angle of elevation from his head becomes 60° from 45°. How much distance (in metres) Mohit covered towards the building?
19.4 (3 - √3)
Effective height above Mohit's head = \(60-1.8=58.2\) m.
Initial position (45°): \(\tan 45^\circ=\tfrac{58.2}{x_1}\Rightarrow x_1=58.2\) m.
Final position (60°): \(\tan 60^\circ=\tfrac{58.2}{x_2}\Rightarrow x_2=\tfrac{58.2}{\sqrt 3}\) m.
Distance walked: \(x_1-x_2=58.2\left(1-\tfrac{1}{\sqrt 3}\right)=\dfrac{58.2(3-\sqrt 3)}{3}=\mathbf{19.4(3-\sqrt 3)}\) m.
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