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Question

Question 6
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.


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Solution



Let the boy initially stands at point Y with inclination 30 and then he approaches the building to the point X with inclination 60.
XY is the distance he walked towards the building.
Also, XY = CD
Height of the building = AZ = 30 m
AB=AZBZ=(301.5)=28.5m
As per the question,
In right ΔABD,
tan 30=ABBD
13=28.5BD
BD=28.53m
Also,
In right ΔABC,
tan 60=ABBC
3=28.5BC
BC=28.53=28.533 m
XY=CD=BDBC =(28.5328.533) =28.53(113) =28.53×23 =573 =193 m
Thus, the distance boy walked towards the building is 193 m.

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