Apoorv is standing in the centre of a rectangular park and observes that the angle of elevation of the top of a lamp post at a corner of the park in `60^(@)`. He then moves diagonally towards the opposite corner of the park and observes that the angle of elevation is now `beta`, then the value of `beta` is

Options

A. `45^(@)`
B. `30^(@)`
C. `tan^(-1) (sqrt(3)//2)`
D. `tan^(-1) (2//sqrt(3))`

Answer

To solve the problem, we need to analyze the situation with respect to the geometry of the rectangular park and the angles of elevation observed by Apoorv.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Apoorv is standing at the center of a rectangular park.
- There is a lamp post at one corner of the park, and the height of the lamp post is denoted as \( h \).
- The angle of elevation from the center of the park to the top of the lamp post is given as \( 60^\circ \).

2. **Positioning the Points**:
- Let the corners of the rectangular park be labeled as \( C \) (where the lamp post is), \( D \), \( A \) (the opposite corner where Apoorv moves), and \( B \) (the center of the park).
- The distance from the center \( B \) to the corner \( C \) is \( x \).

3. **Using the Tangent Function**:
- From triangle \( C B D \), we can use the tangent of the angle of elevation:
\[ \tan(60^\circ) = \frac{h}{x} \]
- Since \( \tan(60^\circ) = \sqrt{3} \), we can write:
\[ \sqrt{3} = \frac{h}{x} \quad \Rightarrow \quad h = x \sqrt{3} \quad \text{(Equation 1)} \]

4. **Moving to the Opposite Corner**:
- Apoorv then moves to corner \( A \). The distance from \( A \) to the lamp post at \( C \) is now \( 2x \) (since he moved diagonally across the rectangle).
- The angle of elevation from point \( A \) to the top of the lamp post is \( \beta \).

5. **Applying the Tangent Function Again**:
- From triangle \( A C D \), we can express the tangent of the new angle of elevation:
\[ \tan(\beta) = \frac{h}{2x} \]
- Substituting \( h \) from Equation 1:
\[ \tan(\beta) = \frac{x \sqrt{3}}{2x} = \frac{\sqrt{3}}{2} \quad \text{(Equation 2)} \]

6. **Finding the Angle \( \beta \)**:
- We know that:
\[ \tan(\beta) = \frac{\sqrt{3}}{2} \]
- To find \( \beta \), we take the inverse tangent:
\[ \beta = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

### Final Answer:
Thus, the value of \( \beta \) is:
\[ \beta = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \]