Question

An aeroplane is flying at a constant speed, parallel to the horizontal ground at a height of 5 kms. A person on the ground observed that the angle of elevation of the plane is changed from \(15^\circ\) to \(30^\circ\) in the duration of 50 seconds, then the speed of the plane (in kmph) is

• A. 100
• B. 720
• C. 360
• D. 540 Correct Answer

Hint:

For height and distance problems, draw a clear diagram. Use trigonometric ratios (tan, sin, cos) to relate angles, heights, and distances. Be careful with units and ensure they are consistent before calculating speed (distance/time). Values: \( \tan 15^\circ = 2-\sqrt{3} \), \( \tan 30^\circ = 1/\sqrt{3} \), \( \tan 45^\circ = 1 \), \( \tan 60^\circ = \sqrt{3} \). 1 hour = 3600 seconds.

Answer 1

The Correct Option is B

Step 1: Draw a diagram.
Let P be the position of the observer on the ground.
Let A and B be the two positions of the aeroplane.
The height of the aeroplane is \( h = 5 \) km.
Let AC and BD be perpendiculars from A and B to the ground.
So AC = BD = 5 km.
Angle of elevation at A is \( \angle CPA = 15^\circ \).
Angle of elevation at B is \( \angle DPB = 30^\circ \).
The observer is at P.
So C and D are points on the ground below A and B.
Since the plane flies parallel to the ground, CD is the horizontal distance covered.
Let PC = \(x_1\) and PD = \(x_2\).
The distance covered by the plane is \( d = CD = |x_1 - x_2| \).
The plane moves from the smaller angle of elevation to the larger, so it's moving towards the observer.
Let's assume the observer is P.
Initial position A, final position B.
So, PC = \(x_1\), PB' = \(x_2\) where B' is the point on ground below B.
In \( \triangle PCA \), \( \tan 15^\circ = \frac{AC}{PC} = \frac{5}{x_1} \implies x_1 = \frac{5}{\tan 15^\circ} \).
In \( \triangle PDB \), \( \tan 30^\circ = \frac{BD}{PD} = \frac{5}{x_2} \implies x_2 = \frac{5}{\tan 30^\circ} \).
The distance covered by the plane is \( d = x_1 - x_2 \).

Step 2: Calculate \( \tan 15^\circ \) and \( \tan 30^\circ \).
\( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
\( \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1+\tan 45^\circ \tan 30^\circ} = \frac{1 - 1/\sqrt{3}}{1+1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \).
Rationalize \( \tan 15^\circ \): \( \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3+1-2\sqrt{3}}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3} \).

Step 3: Calculate \(x_1\) and \(x_2\).
\( x_1 = \frac{5}{2-\sqrt{3}} = \frac{5(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{5(2+\sqrt{3})}{4-3} = 5(2+\sqrt{3}) = 10+5\sqrt{3} \) km.
\( x_2 = \frac{5}{1/\sqrt{3}} = 5\sqrt{3} \) km.

Step 4: Calculate the distance \(d\) covered by the plane.
\( d = x_1 - x_2 = (10+5\sqrt{3}) - 5\sqrt{3} = 10 \) km.

Step 5: Calculate the speed of the plane.
The time taken is \( t = 50 \) seconds.
We need speed in kmph.
Convert time to hours: \( 50 \text{ seconds} = \frac{50}{3600} \text{ hours} = \frac{5}{360} \text{ hours} = \frac{1}{72} \text{ hours} \).
Speed \( v = \frac{\text{distance}}{\text{time}} = \frac{d}{t} = \frac{10 \text{ km}}{1/72 \text{ hours}} \).
\[ v = 10 \times 72 = 720 \text{ kmph} \] This matches option (2).