Question

an aeroplane flying horizontally 2km above the ground is observed at an angle 60 degree at a point on the ground if after 10 seconds the elevation is observed to be 30 degree the unifrome speed in km/hour of the aeroplane is​

Answer 1

Answer:

Let O be the point of observation and A be the position of the aeroplane such that ∠AOC=60

∘

and AC=1km

After 10seconds, let B be the position of the aeroplane such that ∠BOD=30

∘

and BD=1km

In right angled triangle,△AOC,

tan60

∘

=

OC

AC

⇒

3

=

OC

1

⇒OC=

3

1

In right angled triangle,△ODB,

tan30

∘

=

OD

BD

⇒

3

1

=

OD

1

⇒OD=

3

Now, CD=OD−OC=

3

−

3

1

=

3

3−1

=

3

2

Distance covered by the aeroplane in 10 seconds=

3

2

km

Time taken=10secs=

3600

10

=

360

1

hrs

Speed of the aeroplane=

Time

Distance

=

360

1

3

2

=

3

720

3

=240×1.732=415.68km/hr

solution

Answer verified by branily

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