Question

An aeroplane is flying horizontally along a straight line at a height of 3000 m from
the ground at a speed of 160 m/s. Find the time it would take for the angle of elevation
of the plane as seen from a particular point on the ground to change from 60⁰ to 45⁰.
Give your answer correct to the nearest second.
with the figure

Answer 1

Answer:

8 sec

Step-by-step explanation:

tan 45= 1

=> EC= 3000m

tan 60= √3

=> BC = 3000/√3

= 1000√3

EC-BC = EB

= 3000-1000√3

= 1268

if 160m in 1 sec

then 1268 m in 1268/160 sec

7.9 sec

= 8 second.

Answer 2

The time taken by the aeroplane for the angle of elevation of the plane as seen from 60° to 45° is 7.924s.

An aeroplane is flying horizontally along a straight line at a height of 3000 m from the ground at a speed of 160 m/s.

We have to find the time taken for the angle of elevation of the plane as seen from a particular point on the ground to change from 60° to 45°.

Let point of observation from the initial position of aeroplane is x cm.

case 1 : angle of elevation, Ф = 60°

∴ tan60° = height of aeroplane/distance of point of observation to the position of plane.

⇒ √3 = 3000/x

⇒ x = 1000√3 m  ....(1)

case 2 : angle of elevation becomes Ф = 45° , and distance between point of observation and position of plane is (x + 160t) m. where, t is the time taken by aeroplane.

∴ tan45° = height of aeroplane/distance of point to the position of aeroplane.

⇒ 1 = 3000/(x + 160t)

⇒ 3000 = x + 160t

⇒ 3000 = 1000√3 + 160t

⇒ t = (3000 - 1000√3)/160 = 7.924s

Therefore the time taken by the aeroplane for the angle of elevation of the plane as seen from 60° to 45° is 7.924s.

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