Question

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

Answer 1

[ Figure is in the attachment]

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.

Answer 2

Answer:

693.84kmph

Step-by-step explanation:

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

Read more on Brainly.in - https://brainly.in/question/1887709#readmoreLet 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