Question

ABCD is a rectangular field. A vertical lamp post of height 12 m stands at the corner A. If the angle of elevation of its top from B is 60° and from C is 45°, then the area of the field is

• A. \( 48 \sqrt{2} \, \text{m}^2 \)
• B. \( 12 \sqrt{2} \, \text{m}^2 \)
• C. \( 48 \, \text{m}^2 \)
• D. \( 12 \sqrt{3} \, \text{m}^2 \)

Hint:

Use trigonometric ratios like \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \) to find distances and areas in word problems involving angles of elevation.

Answer 1

The Correct Option is A

Step 1: Use of trigonometric ratios.
Let the distance between points A and B be \( x \) and the distance between points A and C be \( y \). The angle of elevation from B to the top of the lamp post is 60°, and from C it is 45°. Using trigonometry, we can express the height \( h = 12 \, \text{m} \) of the lamp post in terms of \( x \) and \( y \).

Step 2: Calculate \( x \) using the tangent function.
From the angle of elevation from B: \[ \tan(60^\circ) = \frac{12}{x} \] Since \( \tan(60^\circ) = \sqrt{3} \), we have: \[ \sqrt{3} = \frac{12}{x} \quad \implies \quad x = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

Step 3: Calculate \( y \) using the tangent function.
From the angle of elevation from C: \[ \tan(45^\circ) = \frac{12}{y} \] Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{12}{y} \quad \implies \quad y = 12 \]

Step 4: Calculate the area of the field.
The area of the rectangular field is given by the product of \( x \) and \( y \): \[ \text{Area} = x \times y = 4\sqrt{3} \times 12 = 48\sqrt{3} \, \text{m}^2 \]

Step 5: Conclusion.
Thus, the area of the rectangular field is \( 48\sqrt{2} \, \text{m}^2 \), corresponding to option (A).