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The angle of elevation of an airplane from a point A on the ground is 60° after a flight of 30 seconds, the angle of elevation changes to $30 °$ . If the plane is flying at a constant height of $36003 m$ , what will be the speed, in km/hour, of the plane?

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860 k m / h r

854 k m / h r

864 k m / h r

874 k m / h r

answer is C.

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Detailed Solution

Given that,
Airplane height =

36003

m
The angle of elevation changes from

60 °

30 °

in 30 seconds.
Question Image

Considering

x

as the linear distance between point A to the plane in the ground.
Substituting

x

for base and

60 °

for

θ

in the equation

tan θ = height base

⇒ tan 60 ° = 36003 x ⇒ 3 = 36003 x [∵ tan 60 ° = 3] ⇒ x = 360033 ⇒ x = 3600 m

Assuming the plane cover

y

distance in 30 seconds then,

x + y

will be the total linear distance between point A and the plane in the ground.
Substituting the value

x + y

for base and

θ

30 °

in the equation

tan θ = height base

, we get,

⇒ tan 30 ° = 36003 x + y ⇒ 13 = 36003 x + y [∵ tan 30 ° = 13] ⇒ x + y = 36003 ⋅ 3 ⇒ x + y = 10800 m

Substituting

x

=3600 in the equation

x + y = 10800

⇒ 3600 + y = 10800 ⇒ y = 10800 − 3600 ⇒ y = 7200

Substituting distance=7200 and time= 30 in the equation

Speed= distance time

⇒ Speed= 720030 ⇒ Speed = 240 m/s ⇒ Speed = 240 × 36001000 km/hr [∵ 1km=1000m, 1hr=3600s] ⇒ S p e e d = 864 km/hr

Thus the speed of aeroplane will be

864 k m / h r

Therefore the correct option is 3.

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