A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Question

A ladder 14 m long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, then the height of the wall is:

Answer 1

The boy was initially at point S. He walked towards the building and reached point T. It is observed that

PR = PQ − RQ = (30 − 1.5) m = 28.5 m = $ \frac{57}{2} $m

In ∆PAR,

$ \frac{PR}{ AR} = \tan 30° $

$ \frac{57}{ 2AR} = \frac{1}{\sqrt3} $

$ AR = (\frac{57}{2 \sqrt3})m $

In ∆PRB,

$ \frac{PR}{ BR} = \tan60° $

$ \frac{57}{ 2BR} = \sqrt 3 $

$ BR = \frac{57}{2\sqrt3} = (\frac{19\sqrt3}{2})\,m $

ST = AB = AR- BR = $ (\frac{57/\sqrt3}{2} - \frac{19\sqrt3}{2} )\,m $
= $ (\frac{38\sqrt3}2 ) \,m $ = $ 19\sqrt3 \,m $

Therefore, he walked $ 19\sqrt3 \,m $ towards the building.

Solution and Explanation