Question

the angle of elevation of an aeroplane from a point on the ground is 60 degree after a flight of 30 seconds the angle of elevation becomes 30 degree if the aeroplane is flying at a constant height of 3000√3M then find the speed of the aeroplane​

Answer 1

200m/s is your answer.

First of all let y be the base when angle is 60° and height =3000√3

then tan 60°= 3000√3/y

and y= 3000

Now let X be the distance covered by the plane on 30 second .

then tan 30°= 3000√3/(x+y)

x+y=9000

x= 6000

so x is the distance it travels in 30 sec .

then speed of the aeroplane =6000/30=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 ✌️