At a place the true value of angle of dip is `60^(@)`. If dip circle is rotated by `phi^(@)` from magnetic meridian, the angle of dip is found to be `tan^(-1)(2)`. Then the value of `phi` is

Options

A. `45^(@)`
B. `15^(@)`
C. `60^(@)`
D. `30^(@)`

Answer

To solve the problem, we need to find the value of \( \phi \) given the true angle of dip \( \phi = 60^\circ \) and the apparent angle of dip \( \phi' = \tan^{-1}(2) \) when the dip circle is rotated by \( \phi \) from the magnetic meridian.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The relationship between the true angle of dip \( \phi \), the apparent angle of dip \( \phi' \), and the angle \( \beta \) (the angle made by the vertical plane with the magnetic meridian) is given by:
\[ \tan \phi' = \frac{\tan \phi}{\cos \beta} \]

2. **Substituting Known Values**:
We know:
- \( \phi = 60^\circ \)
- \( \phi' = \tan^{-1}(2) \)

Therefore, we can express \( \tan \phi \) and \( \tan \phi' \):
\[ \tan \phi = \tan(60^\circ) = \sqrt{3} \]
\[ \tan \phi' = 2 \]

3. **Setting Up the Equation**:
Substitute these values into the relationship:
\[ 2 = \frac{\sqrt{3}}{\cos \beta} \]

4. **Solving for \( \cos \beta \)**:
Rearranging the equation gives:
\[ \cos \beta = \frac{\sqrt{3}}{2} \]

5. **Finding \( \beta \)**:
The angle \( \beta \) can be found using the cosine value:
\[ \beta = 30^\circ \]

6. **Finding \( \phi \)**:
Now, we can find \( \phi \) using the relationship that we established. Since \( \beta \) is the angle made by the vertical plane with the magnetic meridian, we can conclude that:
\[ \phi = 60^\circ \]

### Final Answer:
Thus, the value of \( \phi \) is \( 60^\circ \).