A 1.5m tall boy is standing at some distance from a 30m 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.

Asked by Abhisek | 1 year ago |  96

##### Solution :-

Let the boy initially stand at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.

To Find: The distance boy walked towards the building i.e. XY

From figure,

XY = CD.

Height of the building = AZ = 30 m.

AB = AZ – BZ = 30 – 1.5 = 28.5

Measure of AB is 28.5 m

In right ΔABD,

tan 30° = $$\dfrac{AB}{BD}$$

$$\dfrac{1}{\sqrt{3}}$$ = $$\dfrac{28.5}{BD}$$

BD = $$28.5\sqrt{3}$$ m

Again,

In right ΔABC,

tan 60° = $$\dfrac{AB}{BC}$$

$$\sqrt{3}$$ =$$\dfrac{28.5}{BC}$$

BC = $$28.5\sqrt{3}$$$$\dfrac{28.5\sqrt{3}}{3}$$

Therefore, the length of BC is $$\dfrac{28.5\sqrt{3}}{3}$$ m.

XY = CD = BD – BC = ($$28.5\sqrt{3}-\dfrac{28.5\sqrt{3}}{3}$$

= $$28.5\sqrt{3}$$($$1-\dfrac{1}{3}$$) = $$28.5\sqrt{3}$$ $$\times \dfrac{2}{3}$$

$$57\sqrt{3}$$$$19\sqrt{3}$$  m.

Thus, the distance boy walked towards the building is $$19\sqrt{3}$$ m.

Answered by Pragya Singh | 1 year ago

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