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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.

Answer:

**Solution:**

A common notion in trigonometry, specifically, is the angle of elevation, which has to do with height and distance. It is described as an angle formed by the horizontal plane and an oblique line between the observer's eye and a target above it.

Given,

The height of the tall boy (AS) = 1.5 m

The length of the building (PQ) = 30 m

Let the initial position of the boy be S. And, then he walks towards the building and reached at the point T.

From the fig. we have

AS = BT = RQ = 1.5 m

PR = PQ – RQ = (30 – 1.5)m = 28.5 m

In ΔPAR,

tan 30^{o} = PR/AR

1/ √3 = 28.5/ AR

AR = 28.5√3

In ΔPRB,

tan 60^{o} = PR/BR

√3 = 28.5/ BR

BR = 28.5/√3 = 9.5√3

So, ST = AB = AR – BR = 28.5√3 – 9.5√3 = 19√3

**Therefore, the distance which the boy walked towards the building is 19√3 m.**

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