The angle of elevation of an aeroplane flying in a horizontal straight line from a fixed point at four successive observations are alpha beta gamma and Delta the observations been taken at equal intervals of time is you mean that the speed of the aeroplane is uniform show that 3 cos squared theta minus cot square equals to cos square alpha minus cot square delta
Answers
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.