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.
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Answer:
19√3m
explanation:
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
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