Question

The angle of elevation if an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If aeroplane is flying at a constant height of 1500√3 m, find speed in km/hr

Answer 1

A is the position of the observer.
segment BC is the height at which the jet is flying. BC = ED = 1500√3 m
segment BE is the time taken by the jet which is 15 seconds.
Angle BAC and EAD are the angles of elevation.
BAC = 60° and EAD = 30°

In right angled ∆ BAC,
tan 60° = BC/AC .......( side opp to 60°)
√3 = 1500√3/AC
AC = 1500 m

In right angled ∆ EAD,
tan 30° = ED/AD .......( side opp to 30°)
1/√3 = 1500√3/AD
AD = 1500×√3×√3
AD = 1500×3
AD = 4500 m

AC + CD = AD ........( since A - C - D)
1500 + CD = 4500
CD = 4500 - 1500
CD = 3000 m

BCDE is a rectangle ....( opp side are
congruent)
CD = BE = 3000 m
distance covered in 15 seconds = 3000m
Speed = distance/time
= BE / 15 sec
= 3000 m / 15 sec
= (3000/1000)÷(15/3600).....
.....[1km=1000m and 1hr=3600sec]
= (3000/1000)×(3600/15)
= 3 × 3600/15
= 3600/5
= 720 km/hr
The speed of the jet is 720 km/hr.
I hope this would have helped.
please mark it as a brainliest answer.
The figure is attached as a pic.

Answer 2

Answer:

Step-by-step explanation: