the angle of elevation of a cloud from a point 1.5km above lake is 60* and angle of depression of reflection is 30*. find the height of cloud above the lake
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Answer:
Let P and Q be the two position of the plane.
Given angles of elevation of the plane in two position P and Q as ∠PAB=60
0
and ∠QAC=30
0
PB=QC= 15 km
Now in right angled △ABP
tan60
0
=
AB
BP
⇒
3
=
AB
1.5
⇒AB=
3
1.5
⇒(0.5)
3
km
Again in right angled triangle △ACQ,
tan30
0
=
AC
QC
⇒
3
1
=
AB
1.5
⇒AC=(1.5)
3
km
PQ=BC=AC−AB
=(1.5)
3
−(0.5)
3
=(1.5−0.5)
3
=
3
km
The plane travels PQ=
3
km in 15 secs
∴ Speed of the aeroplane =
time
distance
⇒
3600
15
3
⇒240
3
km/h
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Answer:
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