Question
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° from 60°, as he walks towards the building. Find the distance he walked towards the building.
Solution
Let AC be the observer of height 1.5 m and BE be the building of height 30 m. The angle of elevation from the eyes of observer increases from 30° to 60° to the top of the building.
Fig. 9.8.
i.e., ∠ECD = 30° and ∠EFD = 60°.
Let CF = x m and FD = y m
In right triangle EDF, we have
In right triangle EDC, we have
Comparing (i) and (ii), we get
Hence, distance (CF) 
