Question

The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30". If the aeroplane is flying at a constant height of 3000V3 m, find the speed of the aeroplane.​

Answer 1

Answer:

tan60

∘

=

x

3000

3

x=

3

3000

3

=3000m

tan30

∘

=

1/

3

3000

3

=9000m

distance covered = 9000- 3000

= 6000 m.

speed =

30

6000

=200m/s

Answer 2

Answer:

${ \purple{ \tan 60° = \frac{3000 \sqrt{3} }{x} }}$

${ \purple{x= \frac{3000 \sqrt{3} }{ \sqrt{3} } = 3000m}}$

${ \purple{ \tan 30° = \frac{3000 \sqrt{3} }{ \frac{1}{ \sqrt{3} } } = 9000m}}$

Distance covered :

${ \purple{ 9000 - 3000 = 6000m}}$

${ \purple{speed = \frac{6000}{30} }}$

${ \huge{ \boxed{ \purple{speed = 200 m/s}}}}$

Step-by-step explanation:

Hope this helps you ✌️