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 solve the problem, we must find the distance between two points on the ground, given the angles of depression from a 50 m tall building. We will use trigonometry to determine the horizontal distances from the building to each point. 

Let's assume:

• Point A is the point with a \(45^\circ\) angle of depression.
• Point B is the point with a \(30^\circ\) angle of depression.

1. For Point A, with an angle of depression of \(45^\circ\), the horizontal distance \(d_1\) from the building to Point A can be calculated using the tangent function: \(\tan(45^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = 1\). Thus, the equation is:
   \(\text{d}_1 = h \cdot \tan(45^\circ) = 50 \times 1 = 50 \, \text{m}\).
2. For Point B, with an angle of depression of \(30^\circ\), the horizontal distance \(d_2\) to Point B is determined by: \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), giving us the equation:
   \(\text{d}_2 = h \cdot \tan(30^\circ) = 50 \times \frac{1}{\sqrt{3}} = \frac{50\sqrt{3}}{3} \, \text{m}\).
3. The total distance between the two points on the ground is the difference between these distances:
   \(\Delta d = \text{d}_2 - \text{d}_1 = \frac{50\sqrt{3}}{3} - 50\).
4. Simplifying this, we compute:
   \(\Delta d = \frac{50\sqrt{3} - 150}{3} = \frac{150(\sqrt{3} - 1)}{3} = 50(\sqrt{3} - 1) \, \text{m}\).

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

Answer 2

To solve this problem, we need to find the horizontal distance between two points on the ground from which the angles of depression \(45^\circ\) and \(30^\circ\) are observed from the top of a 50 m tall building.

We will use trigonometry to solve this. Let's denote:

• \(A\) as the top of the building.
• \(B\) as the point on the ground where the angle of depression is \(45^\circ\).
• \(C\) as the point on the ground where the angle of depression is \(30^\circ\).
• \(H = 50 \, \text{m}\) as the height of the building.
• \(AB = x \) as the distance from the base of the building to point \(B\).
• \(AC = y\) as the distance from the base of the building to point \(C\).

Step 1: Calculate \(x\) using \(45^\circ\) angle

For a \(45^\circ\) angle of depression, we use the tangent function:

\(\tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{H}{x}\)

Since \(\tan(45^\circ) = 1\), we have:

\(1 = \frac{50}{x} \Rightarrow x = 50 \text{ m}\)

Step 2: Calculate \(y\) using \(30^\circ\) angle

For a \(30^\circ\) angle of depression, we use the tangent function:

\(\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{H}{y}\)

Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we have:

\(\frac{1}{\sqrt{3}} = \frac{50}{y} \Rightarrow y = 50\sqrt{3} \text{ m}\)

Step 3: Find \(BC\), the distance between points \(B\) and \(C\)

The distance \(BC\) is the difference between \(AC\) and \(AB\):

\(BC = y - x = 50\sqrt{3} - 50\)

Factoring out 50 gives:

\(BC = 50(\sqrt{3} - 1) \text{ m}\)

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