Q. 224.3( 25 Votes )

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.

Answer :

The height of the boy is DC = 1.5 m
Let he was at point C initially and then moved to point F.
Let CF = x 
Now DC is parallel to PB.
DP is parallel to CB.
⇒ DE = CF
 EP = FB
Now, 

AP = AB - BP

 = 30-1.5
28.5 m.

Since the tower is vertical,
∠APE = 90°
We know, in a right-angled triangle,

In ∆ADP,




DP = 28.5√3  ------------(1)


In ∆AEB,



 



DE = DP - EP





19√3 m.


Therefore the walking distance of boy is 19√3 m.

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