Question

From an airplane flying, vertically above a horizontal road, the angles of depression of two consecutive stones on the same side of the airplane are observed to be 30° and 60° respectively. The height at which the airplane is flying in km is :

• (A) $\frac{4}{\sqrt{3}}$
• (B) $\frac{\sqrt{3}}{2}$
• (C) $\frac{2}{\sqrt{3}}$
• (D) $2$

Answer 1

The Correct Option is B

 To solve this problem, we need to determine the height at which the airplane is flying based on the angles of depression observed from it. The problem states that the airplane observes angles of depression of 30° and 60° to two consecutive stones on the same side.

Let's denote the following:

• h: the height of the airplane above the road (which we need to find).
• x: the horizontal distance from the point directly below the airplane to the first stone.

The angle of depression of 60° to the first stone means the angle with the vertical (angle of elevation) from the stone to the airplane is 60°. Similarly, for the second stone with an angle of depression of 30°, the angle of elevation is 30°.

Using trigonometry, we have:

1. For the first stone:

\[\tan(60°) = \frac{h}{x}\]

From this equation, we can express h as:

\[h = x \cdot \sqrt{3}\]

2. For the second stone, which is farther away than the first stone at a distance (x + d):

\[\tan(30°) = \frac{h}{x + d}\]

Rewriting, we get:

\[h = (x + d) \cdot \frac{1}{\sqrt{3}}\]

Since both expressions equal h, we can equate them to find d:

\[x \cdot \sqrt{3} = (x + d) \cdot \frac{1}{\sqrt{3}}\]

Solving this equation:

\[x \cdot \sqrt{3} \cdot \sqrt{3} = x + d\]

The equation becomes:

\[3x = x + d\]

Solving for d:

\[d = 2x\]

Substitute d = 2x back in h using the expression from the second stone:

\[h = (x + 2x) \cdot \frac{1}{\sqrt{3}} = 3x \cdot \frac{1}{\sqrt{3}}\]

Thus, the height is:

\[h = x \cdot \sqrt{3}\]

But then we have:

\[\Rightarrow h = \frac{\sqrt{3}}{2} \text{ km}\]

Thus, the correct answer is:

(B) \(\frac{\sqrt{3}}{2}\)