Question

an aeroplane flying at the height of 9000m from the ground passes vertically above another aeroplane at an instant , when the angle of elevation of the two planes from the same point on the ground are 60 and 45 degree respertively. find the vertical distance between the aeroplanes at that instant.

Answer 1

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Answer 2

Answer:

vertical distance between the aeroplanes  = 9000- 5202.3 = 3797.7 m

Step-by-step explanation:

given, plane A is at a height of 9000 m

angle of elevation between plane A and ground is  60 °

angle of elevation between plane B and ground is 45°

to find: vertical distance between the planes

let h be the vertical distance between plane B and ground

refer to the image attached

< DAC = 45° & < DAB = 60°

DB = 9000 m & CD = 'h' m

tan 45 = $\frac{h}{AD}$

1 = $\frac{h}{AD}$

AD = h ____eq(1)

tan 60 = $\frac{9000}{AD}$

√3 = $\frac{9000}{AD}$

AD = $\frac{9000}{\sqrt{3} }$ _____eq(2)

compare (1) & (2)

h =  $\frac{9000}{\sqrt{3} }$  = 5202.3 m

vertical distance between plane B and ground is 5202.3 m

and vertical distance between plane A and ground is 9000 m

vertical distance between the aeroplanes  = 9000- 5202.3 = 3797.7 m