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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

**Solution:**

Let's represent the situation using a diagram according to the given question.

Distance walked towards the building RQ = PR - PQ

Trigonometric ratio involving AP, PR and ∠R and AP, PQ and ∠Q is tan θ [Refer the diagram to visualise AP, PR and PQ]

In ΔAPR

tan R = AP/PR

tan 30° = 28.5/PR

1/√3 = 28.5/PR

PR = 28.5 × √3 m

In ΔAPQ

tan Q = AP/PQ

tan 60° = 28.5/PQ

√3 = 28.5/PQ

PQ = 28.5 / √3 m

Therefore,

PR - PQ = 28.5√3 - 28.5/√3

= 28.5 (√3 - 1/√3)

= 28.5 ((3 - 1)/√3)

= 28.5 (2/√3)

= 57/√3

= (57 × √3)/(√3 × √3)

= (57√3)/3

= 19√3 m

The distance walked by the boy towards the building is 19√3 m.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 9

**Video Solution:**

## 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

Maths NCERT Solutions Class 10 Chapter 9 Exercise 9.1 Question 6

**Summary:**

If 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, then the distance he walked towards the building is 19√3 m.

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