Question

From the top of a 60 m high building, the angles of depression to two points on the ground on the same side of the building are \( 30^\circ \) and \( 60^\circ \). What is the distance between the two points?

• A. \( 40\sqrt{3} \, \text{m} \)
• B. \( 20\sqrt{3} \, \text{m} \)
• C. \( 60 \, \text{m} \)
• D. \( 80 \, \text{m} \)

Hint:

Use trigonometric ratios to find distances in height and distance problems.

Answer 1

The Correct Option is A

To find the distance between the two points on the ground, we use trigonometry involving angles of depression. Let's denote the points on the ground as \( A \) and \( B \), with \( A \) being the point closer to the building and \( B \) the further point. Let the distance from the building to point \( A \) be \( x \), and the distance from the building to point \( B \) be \( y \). 

The height of the building is \( h = 60 \, \text{m} \). For the angles of depression, we use the fact that from the top of the building, the angle of depression is equal to the angle of elevation from the ground.

Step 1: Angle of depression to point \( A \)

For point \( A \), the angle of depression is \( 60^\circ \). Let \( \angle Z = 60^\circ \):

\(\tan(60^\circ) = \frac{h}{x}\)
\(\sqrt{3} = \frac{60}{x}\)
\(x = \frac{60}{\sqrt{3}}\)
\(x = 20\sqrt{3} \, \text{m}\)

Step 2: Angle of depression to point \( B \)

For point \( B \), the angle of depression is \( 30^\circ \). Let \( \angle Y = 30^\circ \):

\(\tan(30^\circ) = \frac{h}{y}\)
\(\frac{1}{\sqrt{3}} = \frac{60}{y}\)
\(y = 60\sqrt{3} \, \text{m}\)

Step 3: Find the distance between points \( A \) and \( B \)

The distance between two points \( A \) and \( B \) is \( y - x \):
\(y - x = 60\sqrt{3} - 20\sqrt{3}\)
\(y - x = 40\sqrt{3} \, \text{m}\)

Thus, the distance between points \( A \) and \( B \) on the ground is \( 40\sqrt{3} \, \text{m} \).

Answer 2

To solve this problem, we need to determine the distance between two points on the ground based on the angles of depression from the top of a building. We are given:

• A building height of \(60 \, \text{m}\)
• Angles of depression: \(30^\circ\) and \(60^\circ\)

Let's denote:

• Point \(A\) as the point where the angle of depression is \(60^\circ\)
• Point \(B\) as the point where the angle of depression is \(30^\circ\)
• The foot of the building as point \(C\). Hence, \(AC\) and \(BC\) are horizontal distances from the building to points \(A\) and \(B\), respectively.

We apply trigonometry to calculate these distances. For each position: From the angle \(60^\circ\):

• Using \(\tan(60^\circ)=\sqrt{3}\), we have: \[\tan(60^\circ)=\frac{\text{opposite side}}{\text{adjacent side}}\] \[\sqrt{3}=\frac{60}{AC}\Rightarrow AC=\frac{60}{\sqrt{3}}=20\sqrt{3} \, \text{m}\]

From the angle \(30^\circ\):

• Using \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\), we have: \[\tan(30^\circ)=\frac{\text{opposite side}}{\text{adjacent side}}\] \[\frac{1}{\sqrt{3}}=\frac{60}{BC}\Rightarrow BC=60\sqrt{3} \, \text{m}\]

Finally, the distance \(AB\) between the two points is obtained by: \[AB=BC-AC=60\sqrt{3}-20\sqrt{3}=40\sqrt{3} \, \text{m}\]So, the correct answer is \(40\sqrt{3} \, \text{m}\).