A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. 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.

Given, AF = 30m, DF = BE = 1.5m,
AD = 28.5 m
In triangle ABD
∠ABD = 30°
We know,
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⇒ BD = 28.5 √3
Now, in triangle ACD,
∠ABD = 60°
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Multiplying and dividing the fraction by √3, we get

⇒ CD = 9.5√3
∴ distance he walked towards the building = BD – CD
= 28.5√3 –9.5√3
= 19√3 m
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