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+1 vote
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in Mathematics by (600 points)

An aeroplane flying horizontally at a height of 2500m above ground observed at an elevation of 60 deg. if after 15 secs, the angle of elevation is observed to be 30 deg, find the speed of the aeroplane in km/h

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1 Answer

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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 in 15 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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