The angle of elevation of a jet fighter from a point A on the ground is 600. After a flight of 15 seconds, the angle of elevation changes to 300. If the jet is flying at a speed  of  720  km/hr, find  the  constant  height  at  which  the  jet  is  flying.(Use √3 =1.732                         (Ans: 2598m)

Ans: 36 km / hr = 10m / sec

720 km / h = 10 x 720/36

Speed = 200 m/s

Distance of jet from

AE = speed x time

= 200 x 15

= 3000 m

tan 60^o = AC/BC (opposite side/adjacentside)

√3 = AC/ BC

BC √3 = AC

AC = ED (constant height)

∴ BC √3 = ED ...................1

tan 30^o = ED  /BC + CD (opposite side/adjacentside)

1/√3  =  ED/ BC + 3000

BC + 3000/√3 =ED

BC + 3000/√3 =BC√3 (from equation 1)

BC + 3000 = 3BC

3BC - BC = 3000

2 BC = 3000

BC = 3000/2

BC = 1500 m

ED = BC √3 (from equation 1)

= 1500 √3

= 1500 x 1.732

ED = 2598m

∴ The height of the jet fighter is 2598m.