Question

We have all seen aeroplanes flying in the sky, but we might not have thought of how they actually reached the correct destination. Air Traffic Control (ATC) is a service provided by ground-based air traffic controllers who direct an aircraft on the ground and through a given section of controlled airspace and can provide advisory services to the aircraft in non-controlled airspace. All this air traffic is managed and regulated using various concepts based on coordinate geometry and trigonometry. At a given instance, ATC finds that the angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds, it is observed that the angle of elevation changes to 30°. The height of the plane remains constant as \(3000 \sqrt{3} \mathrm{~m}\).

Based on the above information, answer the following questions.

1. Draw a neat and well-labelled figure to represent the above information.
2. What is the distance travelled by the plane in 30 seconds?
3. Keeping the height constant, during the above flight, it was observed that after \(15(\sqrt{3}-1)\) seconds, the angle of elevation changed to \(45^{\circ}\). How much is the distance travelled in that duration?
4. What is the speed of the plane in km/hr?

Answer 1

i. Figure:

Let A be the point of observation on the ground. Let AX be the horizontal line through A. Let P and Q be the two positions of the plane at angles of elevation \(60^\circ\) and \(30^\circ\) respectively. Let \(PB \perp AX\) and \(QC \perp AX\). Then \(PB = QC = 3000\sqrt{3}\) m.

ii. Distance travelled in 30 seconds

In right \(\triangle ABP\): \[\cot 60^\circ = \frac{AB}{PB}\]

\[\Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{3000\sqrt{3}}\]

\[\Rightarrow AB = 3000 \text{ m}\]

In right \(\triangle ACQ\): \[\cot 30^\circ = \frac{AC}{QC}\]

\[\Rightarrow \sqrt{3} = \frac{AC}{3000\sqrt{3}}\]

\[\Rightarrow AC = 9000 \text{ m}\]

Therefore, \[PQ = BC = AC - AB\] 

\[= 9000 - 3000\]

\[= 6000 \text{ m}\]

iii. Distance travelled in \(15(\sqrt{3}-1)\) seconds when angle is 45°

Let D be the point on the ground beneath the plane when the angle of elevation is \(45^\circ\). Let the plane be at R such that \(DR \perp AX\) and \(DR = 3000\sqrt{3}\) m.

In right \(\triangle ADR\): \[\cot 45^\circ = \frac{AD}{DR}\] 

\[\Rightarrow 1 = \frac{AD}{3000\sqrt{3}}\]

\[\Rightarrow AD = 3000\sqrt{3} \text{ m}\]

Distance travelled PR = \[BD = AD - AB\] 

\[= 3000\sqrt{3} - 3000\] 

\[= 3000(\sqrt{3} - 1)\] m.

iv. Speed of the plane

From (ii), distance 6000 m is covered in 30 s. 

\[\text{Speed} = \frac{6000 \text{ m}}{30 \text{ s}} = 200 \text{ m/s}\].

Convert to km/hr: \[200 \times \frac{18}{5} = 720 \text{ km/hr}\].