Question

the angle of elevation of an aeroplane from a point on the ground is 45 degree after flight of 15 seconds the elevation changes to 30 degree if the aeroplane is flying at a height of 3000 m find the speed of the aeroplane

Answer 1

Let the initial position of the plane be at A and it moves to the next position at D in 15 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 = 45° 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 = 3000 m Tan 45° = AC/BC 1 = 3000/BCBC = 3000 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 = 3000 m Tan 30° = DE/BE 1/√3 = 3000/BE BE = 5196.15 mDistance travelled by the plane = BE – BC = 5196.15 – 3000 = 2196.15 m Time taken to travel = 15 sec Speed of the plane = 2196.15 / 15 = 146.41 m/sec. 

Answer 2

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