an aeroplane, at 2 points A and B flying horizontally at a uniform speed of 20m/sec. it is found that it takes 10 minutes to fly from A to B . the angle of elevation of A and B from a point P on the ground are 30°and 60° respectively
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Answered by
1
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
Answered by
6
Answer:
1.b) 12km
Step-by-step explanation:
1. time = 10min = 10×60 = 600 sec
speed = distance / time
20 = distance/600
distance = 20× 600
distance = 12000m or 12km
Answer is b) 12 km
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