A vertical lamp-post of height 9 metres stands at the corner of a rectangular field. The angle of elevation of its top from the farthest corner is `30^@`, while from another corner it is `45^@`. The area of the field is

Options

A. `81sqrt2m^2`
B. `9sqrt2m^2`
C. `81sqrt3m^2`
D. `9sqrt3m^2`

Explanation

To solve the problem step by step, we will analyze the situation involving the lamp-post, the angles of elevation, and the dimensions of the rectangular field.

### Step 1: Understanding the Setup
We have a vertical lamp-post of height 9 meters at point A, which is one corner of a rectangular field ABCD. The angles of elevation from the farthest corner (C) and another corner (D) are given as 30° and 45°, respectively.

### Step 2: Finding the Length AD
From corner D, the angle of elevation to the top of the lamp-post (point A) is 45°. We can use the tangent function:
\[ \tan(45^\circ) = \frac{\text{height}}{\text{base}} \implies 1 = \frac{9}{AD} \implies AD = 9 \text{ meters} \]

### Step 3: Finding the Length AC
From corner C, the angle of elevation to the top of the lamp-post is 30°. Again, we use the tangent function:
\[ \tan(30^\circ) = \frac{\text{height}}{\text{base}} \implies \frac{1}{\sqrt{3}} = \frac{9}{AC} \implies AC = 9\sqrt{3} \text{ meters} \]

### Step 4: Finding Length DC
Now, we can find the length DC using the Pythagorean theorem in triangle ADC:
\[ AC^2 = AD^2 + DC^2 \]
Substituting the known values:
\[ (9\sqrt{3})^2 = 9^2 + DC^2 \implies 243 = 81 + DC^2 \implies DC^2 = 243 - 81 = 162 \implies DC = \sqrt{162} = 9\sqrt{2} \text{ meters} \]

### Step 5: Finding the Area of the Rectangle
The area of rectangle ABCD can be calculated as:
\[ \text{Area} = AD \times DC = 9 \times 9\sqrt{2} = 81\sqrt{2} \text{ square meters} \]

### Final Answer
The area of the rectangular field is:
\[ \text{Area} = 81\sqrt{2} \text{ square meters} \]

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