Question

An airplane, when 4000m high from the ground, passes vertically above another airplane at an instant when the angles of elevation of the two airplanes from the same point on the ground are 60° and 30°, respectively. Find the vertical distance between the two airplanes:

• A. $\frac{16000}{7}$ m
• B. $\frac{8000}{3}$ m
• C. $\frac{8000}{7}$ m
• D. 1200 m

Hint:

For problems involving angles of elevation and heights, use trigonometric ratios to relate the angles and distances.

Answer 1

The Correct Option is B

Let the point of observation on the ground be O. Let the first airplane be at height 4000m above the ground. Let the second airplane be at height \( h \) meters above the ground. We are given that the angles of elevation of the two airplanes from O are 60° and 30°, respectively. Let the horizontal distance of the first airplane from point O be \( x_1 \), and the horizontal distance of the second airplane from point O be \( x_2 \). Using the tangent of the angles of elevation, we can set up the following relationships: \[ \tan(60^\circ) = \frac{4000}{x_1} \text{and} \tan(30^\circ) = \frac{h}{x_2} \] We know that: \[ \tan(60^\circ) = \sqrt{3}, \tan(30^\circ) = \frac{1}{\sqrt{3}} \] Thus: \[ \sqrt{3} = \frac{4000}{x_1} $\Rightarrow$ x_1 = \frac{4000}{\sqrt{3}} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x_2} $\Rightarrow$ x_2 = \frac{h}{\frac{1}{\sqrt{3}}} = \sqrt{3} h \] Now, since the first airplane is directly above the second airplane, we have: \[ x_1 = x_2 \] Equating the values of \( x_1 \) and \( x_2 \), we get: \[ \frac{4000}{\sqrt{3}} = \sqrt{3} h \] Solving for \( h \), we get: \[ h = \frac{4000}{3} \] Thus, the vertical distance between the two airplanes is: \[ 4000 - h = 4000 - \frac{4000}{3} = \frac{8000}{3} \, \text{m} \]