Question

The angle of elevation of an aeroplane from a point A on the ground is 60° after a flight of 30 sec the angle of elevation changes to to 30° if the plane is flying at a constant height of3600√3m find the speed in Km/hr of the plane

Answer 1

Answer: hey thanks for showing interest..

Step-by-step explanation:

Tan 60 = 3600√3 ÷ AP =√3

3600= ap

Tan 30= 3600√3÷ ab= 1/√3

10800=ab

Pb=10800-3600=7200

Speed =dist./time

10800/30=360m/sec.

But to change it in km/hr

360×1000÷60×60 = 100 km/hr.

Answer 2

Question:

The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 30 sec the angle of elevation changes to to 30°. If the plane is flying at a constant height of 3600√3m, find the speed of the aeroplane.

Answer:

In ΔABE,

tan60=BE/AE

√3=(3600√3)/x

x=3600 m

In ΔACD,

tan30=CD/AD

1/√3=(3600√3)/AD

AD=3600√3*√3

    =10800 m

AD-x=ED

ED=10800-3600

    =7200 m

As Speed=Distance/Time,

    S=7200/30

      =240 m/s