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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° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution
Let the boy stand at point S initially. He walked towards the building and reached at point T.
It can be observed that
PR = PQ - RQ
= (30 - 1.5) m = 28.5 m = `57/2` m
In ΔPAR,
`("PR")/("AR")` = tan 30°
`57/(2"AR") = 1/sqrt3`
`"AR"=57/2sqrt3`
In ΔPRB,
`("PR")/("BR")` = tan 60°
`57/(2"BR") =sqrt3`
`"BR" = 57/(2sqrt3)`
= `((19sqrt3)/2) m`
ST = AB
= AR - BR
= `((57sqrt3)/2 - (19sqrt3)/2)m`
= `((38sqrt3)/2)`m
= `19sqrt3` m
Hence, he walked `19sqrt3` m towards the building.
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