An aeroplane flying horizontally at a height 2500m above the ground is observed at an elevation is of 60 degree and after 15 seconds, the elevation if observed to be 30degre. Find the speed of the aeroplane in Km/hrs.
Answers
Let B & D be the two positions of the plane and let A be the point of observation. Let ABE the horizontal line through A. It is given that angles of elevation of the plane into positions B and D from a point A are 60 °and 30°.
Angle BAC = 60° , angle DAB = 30°
In ∆ ABC
tan 60 °= BC/AC = 2500/AC
√3 = 2500/AC
AC = 2500/√3
IN ∆ AED
tan 30° = ED/AE
1/√3= 2500/AE
AE= 2500√3
BD = CE
CE= AE-AC
BD = 2500√3 - 2500 /√3
BD = 2500 ( √3 - 1/√3)
BD = 2500 ( √3×√3 - 1)/√3
BD = 2500 (3-1)/√3
BD =( 2500 ×2)/√3
BD = 5000 /√3 m
BD =( 5000/ √3 ) × 1/1000 km
BD = 5/√3 km ( Distance)
Plane travels 5/√3 km in15 sec
Time = 15 sec (given )
Time = 15/3600= 1/240 hr
Speed= Distance/time
Speed=( 5/√3) / (1/240)
Speed = 5/√3 × 240
Speed = (5 × 240)/1.732 [ √3= 1.732]
Speed= 1200 /1.732
Speed = 692.84 km/h
Hence, the speed of the aeroplane in km/h is 692.84 km/h.
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