Question

the angle of elevation of an aeroplane from a point on the ground is 60 degree. After a flight of 30 sec. the angle of elevation becomes 30 degree . If the aeroplane is flying at a constant height of 3000√3m, find the speed of the aeroplane.

Answer 1

$1 \div \sqrt{3 } = 3000 \sqrt{3} \div 3000 + x \\ 3000 + x = 3000 \times ( \sqrt{ }3) ^{2}$

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 ✌️