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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 60o = AC/BC (opposite side/adjacentside)
√3 = AC/ BC
BC √3 = AC
AC = ED (constant height)
∴ BC √3 = ED ...................1
tan 30o = 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.
a couple q''s
1/2+1/2
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