In a dip circle , the measurement of an apparent dip is `60^(@)` at a place where true dip is `30^(@)` . If the dip circle is rotated through `90^(@)` then the relating of dip circle is

Options

A. `cot^(-1) (sqrt((8)/(3)))`
B. `cot^(-1)((8)/(3))`
C. `tan^(-1)((8)/(3))`
D. `tan^(-1)(sqrt((8)/(3)))`

Answer

To solve the problem, we will use the relationship between the true dip, apparent dip, and the dip circle's rotation. Let's go through the steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Apparent dip, \( \delta_1 = 60^\circ \)
- True dip, \( \delta = 30^\circ \)

2. **Use the Relationship Between True Dip and Apparent Dip**:
The relationship between the true dip, apparent dip, and the dip circle's angle can be expressed as:
\[ \cot^2 \delta = \cot^2 \delta_1 + \cot^2 \delta_2 \]
where \( \delta_2 \) is the dip measurement after rotating the dip circle.

3. **Calculate \( \cot \) Values**:
- Calculate \( \cot 30^\circ \):
\[ \cot 30^\circ = \frac{1}{\tan 30^\circ} = \sqrt{3} \quad \Rightarrow \quad \cot^2 30^\circ = 3 \]
- Calculate \( \cot 60^\circ \):
\[ \cot 60^\circ = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}} \quad \Rightarrow \quad \cot^2 60^\circ = \frac{1}{3} \]

4. **Substitute the Values into the Equation**:
Substitute \( \cot^2 30^\circ \) and \( \cot^2 60^\circ \) into the equation:
\[ 3 = \frac{1}{3} + \cot^2 \delta_2 \]

5. **Solve for \( \cot^2 \delta_2 \)**:
Rearranging the equation gives:
\[ \cot^2 \delta_2 = 3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3} \]

6. **Find \( \cot \delta_2 \)**:
Taking the square root:
\[ \cot \delta_2 = \sqrt{\frac{8}{3}} = \frac{2\sqrt{2}}{\sqrt{3}} \]

7. **Find \( \delta_2 \)**:
Finally, we find \( \delta_2 \):
\[ \delta_2 = \cot^{-1} \left(\frac{2\sqrt{2}}{\sqrt{3}}\right) \]

### Conclusion:
The relation of the dip circle after rotating through \( 90^\circ \) is:
\[ \delta_2 = \cot^{-1} \left(\sqrt{\frac{8}{3}}\right) \]