Question

WHEN 3000M HIGH AEROPLANES PASSES VERTICALLY ABOVE ANOTHER
AT AN INSTANT WHEN THERE ANGLE OF ELEVATION
AT SAME OBSERVING POINT ARE SAME ELEVATION 60 DEGREE
AND 45 DEGREE HOW MANY METRES LOWER THAN ONE

Answer 1

Answer:

3000-1000√3

Step-by-step explanation:

1.height of first aeroplane is 3000m

2.height of second aeroplane be x m

3.angle of elevation of first aeroplane is 60

4.angle of elevation of second aeroplane is 45.

then for first triangle

tan60= height of first aeroplane/

distance from aeroplane

√3= 3000/distance from plane

distance from aeroplane= 1000√3

in second triangle

tan 45 = height of second plane / distance from aeroplane

1 = x/ 1000√3

1000√3= x

therefore, difference between there height is

height of first aeroplane- height of second aeroplane

3000-1000√3