The angle of elevation of an aeroplane from a point on the ground is 45 degrees.After flying for 15 sec, the angle of elevation changes to 30.If the aeroplane flying at a height of 2500 metres , then speed of the aeroplane
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Let E and D be the two positions of the plane and A be the point of observation.
Let ABC be the horizontal line through A.
It is given that angles of elevation of the plane in two positions E and D are a point A are 60
o
and 30
o
respectively.
∠EAB=60
o
, ∠DAB=30
o
.
It is also given that EB=3000
3
m.
In ΔABE,tan60
o
=
AB
BE
1
3
=
AB
3000
3
AB=3000 m
In ΔACD,tan30
o
=
AC
DC
3
1
=
AC
3000
3
AC=9000 m
Distance =BC=AC–AB=9000–3000=6000 m
The plane travels 6 km in 30 second,
Then the speed of the plane =
30
6000
=200
sec
m
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