Question

An aeroplane when flying at a height of 4000m from ground passes vertically above another plane at an instant when the angles of elevation of two planes from same point on ground are 60degree and 45 degree respectively. find distance between 2 planes

Answer 1

Let A is an aeroplane when it is flying at height 4000m from Ground passes vertical above another Aeroplane B at an instant angle of elevation of two planes from same point D on the ground are 60° and 45° respectively as shown in figure , here distance between two planes is AB = h
And distance between point D and C is x { Let }

Now, from ∆ACD ,
tan60° = perpendicular/base =4000/x
x = 4000/tan60° = 4000/√3 m ---(1)

Again, from ∆BCD ,
tan45° = BC/CD = (4000 - h)/x
1 =(4000 - h)/(4000/√3) [ from equation (1)]
4000/√3 = 4000 - h
h = 4000( 1 - 1/√3) m

Hence, distance between two planes is 1690.59 m

Answer 2

Step-by-step explanation:

Answer:

1690.599 m

Step-by-step explanation:

Refer the attached figure

Height of first Airplane = AB = 4000 m

Height of another plane = BC

The angles of elevation of two planes from the same point on the ground are 60° and 45°. i.e.∠ADB = 60° and ∠CDB = 45°

Let AC be h

CB = AB - AC = 4000-h

In ΔABD

tan \theta = \frac{Perpendicular}{Base}tanθ=BasePerpendicular

tan 60^{\circ} = \frac{AB}{BD}tan60∘=BDAB

\sqrt{3}= \frac{4000}{BD}3=BD4000

BD= \frac{4000}{\sqrt{3}}BD=34000

BD= 2309.401BD=2309.401

In ΔCBD

tan \theta = \frac{Perpendicular}{Base}tanθ=BasePerpendicular

tan 45^{\circ} = \frac{CB}{BD}tan45∘=BDCB

1= \frac{4000-h}{2309.401}1=2309.4014000−h

2309.401=4000-h2309.401=4000−h

h=4000-2309.401h=4000−2309.401

h=1690.599h=1690.599

Hence the vertical distance between the aeroplane at that instant is 1690.599 m