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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