Question

4. 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√3 m, find the speed of the airplane?​

Answer 1

Answer:

in right angle triangle ABE

tanA =BE/AB

tan60 degree =BE/AB

root3=3000root3/AB

AB=3000

IN TRIANGLE

tanA=CD/AC

Tan30degree =CD/AC

1/root3=3000root3/AC

AC=3000root3×root3

=9000m

now AC=AB+BC

9000=3000+BC

BC=9000-3000

=6000

distance travelled by aur in 30s= 6000m

now speep=distance/time

=6000/30

=200m/s

Step-by-step explanation:

i hope it will help you

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 ✌️