Question

The angle of elevation of a jet plane from a point A on the ground is $60^{\circ}$. After a flight of 20 seconds at the speed of $432 km /$ hour, the angle of elevation changes to $30^{\circ} .$ If the jet plane is flying at a constant height, then its height is :

• A. $1800 \sqrt{3} m$
• B. $3600 \sqrt{3} m$
• C. $2400 \sqrt{3} m$
• D. $1200 \sqrt{3} m$

Answer 1

The Correct Option is D

To find the height of the jet plane flying at a constant height, we will use trigonometry. The problem is set up with two angles of elevation, and the plane travels a certain distance in the horizontal direction.

1. First, convert the speed of the jet from km/h to m/s:

\(\text{Speed} = 432 \text{ km/h} = \left( \frac{432 \times 1000}{3600} \right) \text{ m/s} = 120 \text{ m/s}\)

1. Calculate the horizontal distance covered by the plane in 20 seconds at this speed:

\(\text{Distance} = 120 \text{ m/s} \times 20 \text{ s} = 2400 \text{ m}\)

1. Consider the point A from which the angle of elevation is observed. Initially, the angle is \(60^{\circ}\). Let the height of the plane be \(h\) and the initial horizontal distance from point A to the point right under the plane be \(x\). From the angle of elevation:

\(\tan(60^{\circ}) = \frac{h}{x}\)

\(\sqrt{3} = \frac{h}{x}\)

\(h = \sqrt{3}x\)

1. After 20 seconds, the angle changes to \(30^{\circ}\), and the horizontal distance is \(x + 2400 \text{ m}\). Using this:

\(\tan(30^{\circ}) = \frac{h}{x + 2400}\)

\(\frac{1}{\sqrt{3}} = \frac{h}{x + 2400}\)

\(h = \frac{x + 2400}{\sqrt{3}}\)

1. Since both expressions represent the height \(h\), set them equal to solve for \(x\):

\(\sqrt{3}x = \frac{x + 2400}{\sqrt{3}}\)

Multiply through by \(\sqrt{3}\): \(3x = x + 2400\)

\(2x = 2400\)

\(x = 1200 \text{ m}\)

1. Substitute \(x = 1200 \text{ m}\) back into the equation for \(h\):

\(h = \sqrt{3} \times 1200 = 1200\sqrt{3} \text{ m}\)

Therefore, the height of the jet plane is the correct option: \(1200 \sqrt{3} \text{ m}\).