Question

an aeoplane flying horizontally 1000m above the ground is observed at an elevation is 60. After 10 secs, its elevation is observed to be 30,find the speed of aeroplane in km/hr ?

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 $= \frac{1154.6}{10} = 115.46$ m/sec.

Hope This Helps :)