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An aeroplane flying horizontally at a height of 2500 metre above the ground is observe at an elevation of 60degree. If after 15second the angle of elevation is observed to be 30degree find the speed of the aeroplane in km per hour. |
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Answer» 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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