Question

An areoplane is flying a height 2thoushand meter above the ground

Answer 1

Let the initial position of the plane be at A and it moves to the next position at D in 10 sec.

Let BCE be the horizontal line on the ground. B is the point of observation. Two angles of elevation of the plane from the point B are ∠ABC = 60° and ∠DBC = 30°.

ΔABC forms a right angled triangle with AB as the hypotenuse, BC as the base and AC as the height.

AC = height of the plane = 1000 m

Tan 60° = AC/BC

√3 = 1000/BC

BC = 1000 / √3

BC = 1000 / 1.732

BC = 577.4 m

Similarly, ΔDBE forms a right angled triangle with BD as the hypotenuse, BE as the base and DE as the height.

DE = height of the plane = 1000 m

Tan 30° = DE/BE

1/√3 = 1000/BE

BE = √3 x 1000 m

BE = 1.732 x 1000

BE = 1732 m

Distance travelled by the plane = BE – BC = 1732 – 577.4 = 1154.6 m

Time taken to travel = 10 sec

Speed of the plane = 1154.6 / 10 = 115.46 m/sec.

hope this helps....