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
Answer:
Let the boy was standing at point S initially. He walked towards the building and reached at point T.
It can be observed that,
PR=PQ−RQ
⇒ PR=(30−1.5)m=28.5m=
2
57
In △PAR,
AR
PR
=tan30
o
⇒
2AR
57
=
3
1
∴ AR=
2
57
3
m
In △PRB,
BR
PR
=tan60
o
2BR
57
=
3
∴ BR=
2
3
57
=
2
19
3
m
We know, ST=AB
⇒ AB=AR−BR=(
2
57
3
−
2
19
3
)m
∴ AB=
2
38
3
m=19
3
m
Answer:
- BF = 19 √ 3
Step-by-step explanation:
★ GIVEN :
- 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.
★ TO FIND :
- The distance he walked towards the building.
★ SOLUTION :
- In solving this problem firstly we first need to draw a diagram for a clear picture of the scenario. We need to observe the triangles which are formed in the question and then use the formula of tan θ to find the required equations and find the distance the boy travelled.
- Remember to observe the diagram carefully and recall the difference between angle of elevation and angle of depression as many students confuse between them . Also rationalize the answer to remove any radical from the base .
Let the CD to be the Height of building and AB be the Hight of boy
➙ In ∆ CAG ∠ A = 35 °
➙ Tan 30 ° = CG / AG
➙ 1 / √ 3 = 30 - 1.5 / AG
➙ AG = 28 .5 × √ 3 - (i)
Again,
➙ In ∆ CEG , ∠ E = 60°
➙ Tan 60 = CG / EG
➙ √ 3 = 28. 5 / EG
➙ EG = 28.5 / √ 3 × √ 3 / √3 = 28.5 × √3 / 3
➙ EG = 9.5 × √3 - (ii)
➙ AE = BF ∠ 28.5 √3 - 9.5 √3
BF = 19 √3
- Thus we get 19 √ 3 which is the distance he walked towards the building.
