Question

The angle of elevation of a jet plane from a point A on the ground is 60degrees After a flight of
15seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant
height of 1500root over3 meter, find the speed of the jet plane.​

Answer 1

In ΔABC,

tan60 = AC/BC

*3^ = 1500*3^/BC

BC = 1500*3^/*3^

BC = 1500m

In ΔADE,

tan30 = DE/BE

1/*3^ = 1500*3^/BE

1 x BE = 1500*3^ x *3^hi

BE = 1500 x 3

BE = 4500 m

Distance covered by the jet plane in 15 seconds =

BE - BC

=4500 - 1500

= 3000 m

therefore speed of jet plane=

speed = distance/time

speed = 3000/15

speed= 200 m/s

Answer 2

Answer:

200 m/s.

Step-by-step explanation:

Let D and E be the initial and final positions of the plane respectively

(i)

Consider the triangle ABD

tan 60 = BD/AB

√3 = 1500√3/AB

AB = 1500

(ii)

Consider the triangle ACE

tan 30 = CE/AC

1/√3 = 1500√3/AC

AC = 4500

Now,

BC = AC - AB = 4500 - 1500 = 3000 m

DE = BC = 3000 m

i.e., the plan travels a distance of 3000m in 15 seconds.

Therefore, speed of the plane =

= 3000/15

= 200 m/s.

Hope it helps!