Question

From the top of a 50 m tall building, the angles of depression to two points on the ground are \( 45^\circ \) and \( 30^\circ \). Find the distance between the two points.

• A. \( 20(\sqrt{3} - 1) \, \text{m} \)
• B. \( 25(\sqrt{3} - 1) \, \text{m} \)
• C. \( 30(\sqrt{3} - 1) \, \text{m} \)
• D. \( 50(\sqrt{3} - 1) \, \text{m} \)

Hint:

\textbf{Key Fact:} Use tangent of the angle of depression to find horizontal distances from the height.

Answer 1

The Correct Option is D

To determine the distance between two ground points, we utilize the angles of depression from a 50m tall building. Trigonometry will be applied to calculate the horizontal distances from the building to each point.

Assumptions:

• Point A corresponds to a \(45^\circ\) angle of depression.
• Point B corresponds to a \(30^\circ\) angle of depression.

1. For Point A (\(45^\circ\) depression angle), the horizontal distance \(d_1\) from the building is calculated using the tangent function: \(\tan(45^\circ) = 1\). Thus, \(d_1 = h \cdot \tan(45^\circ) = 50 \times 1 = 50 \, \text{m}\).
2. For Point B (\(30^\circ\) depression angle), the horizontal distance \(d_2\) is calculated as: \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). Therefore, \(d_2 = h \cdot \tan(30^\circ) = 50 \times \frac{1}{\sqrt{3}} = \frac{50\sqrt{3}}{3} \, \text{m}\).
3. The distance between the two points is the difference between these horizontal distances: \(\Delta d = d_2 - d_1 = \frac{50\sqrt{3}}{3} - 50\).
4. Simplifying the expression yields: \(\Delta d = \frac{50\sqrt{3} - 150}{3} = \frac{150(\sqrt{3} - 1)}{3} = 50(\sqrt{3} - 1) \, \text{m}\).

The final distance between the two points is \(50(\sqrt{3} - 1) \, \text{m}\).