From the top A of a vertical wall AB of height 30 m, the angles of depression of the top P and bottom Q of a vertical tower PQ are `15^@` and `60^@` respectively, B and Q are on the same horizontal level. If C is a point on AB such that CB = PQ, then the area (in `m^2` ) of the quadrilateral BCPQ is equal to

Options

A. `300 (sqrt3 − 1)`
B. `300 (sqrt3 +1)`
C. `600 (sqrt3 -1)`
D. `200 (sqrt3 -1)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have a vertical wall AB of height 30 m. The angles of depression from point A to the top P and bottom Q of a vertical tower PQ are given as 15° and 60°, respectively. We need to find the area of the quadrilateral BCPQ, where C is a point on AB such that CB = PQ.

### Step 2: Set Up the Diagram
1. Let the height of the tower PQ be \( h \).
2. The height of the wall AB is 30 m.
3. The horizontal distance from A to Q is \( x \).
4. The horizontal distance from A to P is also \( x' \).

### Step 3: Use Trigonometry to Find Distances
Using the angles of depression:
- For angle 15° (to point P):
\[ \tan(15°) = \frac{30 - h}{x'} \]
Rearranging gives:
\[ x' = \frac{30 - h}{\tan(15°)} \]

- For angle 60° (to point Q):
\[ \tan(60°) = \frac{30}{x} \]
Rearranging gives:
\[ x = \frac{30}{\tan(60°)} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \]

### Step 4: Relate Heights and Distances
From the two equations, we have:
- \( x' = \frac{30 - h}{\tan(15°)} \)
- \( x = 10\sqrt{3} \)

### Step 5: Calculate the Height of the Tower PQ
Using the relationship between \( x' \) and \( x \):
\[ \frac{30 - h}{\tan(15°)} = 10\sqrt{3} \]
Substituting \( \tan(15°) \) (which is \( 2 - \sqrt{3} \)):
\[ 30 - h = 10\sqrt{3} \cdot (2 - \sqrt{3}) \]
Calculating the right-hand side:
\[ 30 - h = 20\sqrt{3} - 30 \]
Thus,
\[ h = 30 - (20\sqrt{3} - 30) = 60 - 20\sqrt{3} \]

### Step 6: Find the Area of Quadrilateral BCPQ
The area of quadrilateral BCPQ can be calculated as:
\[ \text{Area} = \text{Base} \times \text{Height} = BC \times PQ \]
Where:
- \( BC = 30 - h = 30 - (60 - 20\sqrt{3}) = 20\sqrt{3} - 30 \)
- \( PQ = h = 60 - 20\sqrt{3} \)

Thus, the area becomes:
\[ \text{Area} = (20\sqrt{3} - 30)(60 - 20\sqrt{3}) \]

### Step 7: Simplify the Area Expression
Expanding the area:
\[ = 1200 - 400\sqrt{3} + 600\sqrt{3} - 600 \]
Combining like terms:
\[ = 600 - 400\sqrt{3} + 600\sqrt{3} = 600 + 200\sqrt{3} \]

### Final Answer
The area of quadrilateral BCPQ is:
\[ \text{Area} = 600( \sqrt{3} - 1) \text{ m}^2 \]