Question

The angle of elevation of an airplane from a point on the ground is 600. After a flight of 30 seconds, the angle of elevation becones 300. If the airplane is flying at a contant hieght of 3000 root 3m, find the speed of the airplane

Answer 1

ANSWER:-

Given:

The angle of elevation of an airplane from a point on the ground is 60°. After a flight of 30 seconds, the angle of elevation becomes 30°. If the airplane is flying at a constant height of 3000√3m.

To find:

Find the speed of the airplane.

Solution:

Let A be the point of observation, C & E be the two points of the plane. It is given that after 30 seconds angle of elevation changes from 60° to 30°.

∠BAC = 60°

∠DAE = 30° &

It is also given that height of the airplane is 3000√3m.

CB= 3000√3m

Since airplane is flying at constant height, therefore, CB= ED = 3000√3m.

In right ∆ABC,

$tan60 \degree = \frac{BC}{AB} \\ \\ = > \sqrt{3} = \frac{3000 \sqrt{3} }{ AB } \\ \\ = > \sqrt{3} AB = 3000 \sqrt{3} \\ \\ = > AB = \frac{3000 \sqrt{3} }{ \sqrt{3} } \\ \\ = > AB = 3000 m .............(1)$

&

In right ∆ADE,

$tan30 \degree = \frac{DE}{AD} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{DE}{AB + BD} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{3000 \sqrt{3} }{AB + BD} \\ \\ = > AB + BD = 3000 \sqrt{3} \times \sqrt{3} \\ \\ = > AB + BD = 3000 \times 3 \\ \\ = > AB + BD = 9000m ...............(2)$

Putting the value of equation (1) in (2), we get;

=) 3000m + BD = 9000m

=) BD= 9000m - 3000m

=) BD= 6000m

Therefore,

Distance travelled in 30 seconds

=) CE = BD= 6000m

Now,

Speed of airplane (m/s)= 6000/30

=) 200m/sec.

&

Speed of airplane (km/h):

=) 200/1000× 3600

=) (2× 360)km/hr.

=) 720km/hr.

Hope it helps ☺️

Answer 2

Answer:

200 m/s

Step-by-step explanation:

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