Two buildings are at a distance of 153√3 from each other. A boy standing at the top of the smaller building is looking at the top and bottom of the other building. The angle of elevation and depression are 60º and 30° respectively. What will be the heights of the two buildings?
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Answer:
The given scenario is shown in the attached figure.
Given:
DE=10
∠CDB=45
o
∠CDA=60
o
To find: AB
Solution:
tan∠CDB=
CD
CB
⇒tan45
o
=
CD
DE
∵CB=DE
1=
CD
10
CD=10 m
tan∠CDA=
CD
CA
⇒tan60
o
=
10
CA
3
=
10
CA
CA=10
3
m
AB=AC+BC
AB=10+10
3
≈27.32 m
Hence, the height of the tower is 27.32 m
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Step-by-step explanation:
Two buildings are at a distance of 15√3 m from each other. A boy standing at the top of
the smaller building is looking at the top and bottom of the other building. The angle of
elevation and depression are 60° and 30° respectively. What will be the heights of the
two buildings?
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