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.

Answer 1

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.