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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. What is the distance he walked towards the building?

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9.5 (33 + 3) m

9.5 (33 − 3) m

9 (33 − 3) m

8.5 (33 − 3) m

answer is B.

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Detailed Solution

Given that,
Angle of elevation in the starting from the boy’s eyes

= 30 o

,
Angle of elevation at the end from the boy’s eyes

= 60 o

,
Height of the boy = 1.5m and,
Height of the building = 30m.
The situation represented diagrammatically is as follows,

Question Image

First, the position of boy is at ‘y’, where the angle of elevation

= 30 o

and as he moves closer to the building i.e., to the position ‘x’, the angle of elevation becomes

60 o

.
Here,
AB = AZ – BZ

⇒

AB = 30 – 1.5 = 28.5m
In

Δ A B D

tan 30 ° = A B B D ⇒ 13 = 28.5 B D ⇒ B D = 28.5 3

Δ A B C,

tan 60 ° = A B B C ⇒ 3 = 28.5 B C ⇒ B C = 28.5 3

The distance travelled by the boy = BD – BC = CD.

⇒ C D = 28.5 3 − 13 ⇒ C D = 28.5 33 − 3 3 ⇒ C D = 9.5 (33 − 3) m

Therefore, the boy travelled

9.5 (33 − 3) m

in the given interval.
The correct option is 2.

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