The horizontal distance between two towers is 120 m. The angle of elevation of the
top and the angle of depression of the bottom of the first tower as observed from the
second are 30 ° and 24 ° respectively. Find the height of the towers. Give your answer
correct to 3 significant figures.
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Answer:
ANSWER
Let AB and CD are two towers.
BD=120m=EC, ∠ACE=30
o
, ∠CBD=24
o
.
In right angle △CBD,
tan24
o
=
BD
CD
=
120
CD
CD=120×0.4452
=53.42m
In right angle △ACE,
tan30
o
=
EC
AE
=
120
AE
AE=120×0.5773
=69.28m
AB=AE+EB
=AE+CD [EB=CD]
=69.28+53.42
=122.7m
Height of 1
st
tower is 122.7m and 2
nd tower is 53.42m
Answered by
0
Let AB and CD are two towers.
BD=120m=EC, ∠ACE=30
o
, ∠CBD=24
o
.
In right angle △CBD,
tan24
o
=
BD
CD
=
120
CD
CD=120×0.4452
=53.42m
In right angle △ACE,
tan30
o
=
EC
AE
=
120
AE
AE=120×0.5773
=69.28m
AB=AE+EB
=AE+CD [EB=CD]
=69.28+53.42
=122.7m
Height of 1
st
tower is 122.7m and 2
nd
tower is 53.42m.
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