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

The distance between two ground points, \( A \) and \( B \), is calculated using trigonometry with angles of depression. Point \( A \) is closer to the building, and point \( B \) is farther. Let \( x \) be the distance from the building to \( A \), and \( y \) be the distance from the building to \( B \).

The building's height is \( h = 60 \, \text{m} \). The angle of depression from the building's top to a point on the ground equals the angle of elevation from that point to the building's top.

Step 1: Calculate distance to point \( A \)

The angle of depression to point \( A \) is \( 60^\circ \). Using the tangent function, where \( \angle Z = 60^\circ \):

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

Step 2: Calculate distance to point \( B \)

The angle of depression to point \( B \) is \( 30^\circ \). Using the tangent function, where \( \angle Y = 30^\circ \):

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

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

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

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