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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. Find the distance he walked towards the building.

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.

Answers (1)

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Given that,
The height of the tall boy (DC) is 1.5 m and the height of the building (AB) is 30 m.
\angle ADF = 30^o and \angle AEF = 60^o

According to the question,
In right triangle AFD,
\\\Rightarrow \tan 30^o=\frac{AF}{DF} = \frac{28.5}{DF}\\\\\Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{DF}
So, DF = (28.5)\sqrt{3}

In right angle triangle \Delta AFE,
\tan 60^o =\frac{AF}{FE}=\frac{28.5}{EF}
\sqrt{3}=\frac{28.5}{EF}
EF = 9.5\sqrt{3}

So, the distance walked by the boy towards the building = DF - EF = 19\sqrt{3}

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manish

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