A dip circle is placed in geographic meridian at a place where dip and declination are known be the respectively `30^(@) and 45^(@).` What dip will be given by dip circle ?

Options

A. `tan ^(-1) ((1)/(sqrt6))`
B. `tan ^(-1) (sqrt ((2)/(3 )))`
C. `tan ^(-1) ((2)/(sqrt3))`
D. `tan ^(-1) ((1)/(2))`

Answer

To find the dip given by a dip circle placed in a geographic meridian where the dip and declination are known, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The **angle of dip (δ)** is the angle between the direction of the Earth's magnetic field and the horizontal plane in the magnetic meridian.
- The **magnetic declination (α)** is the angle between the magnetic meridian and the geographic meridian.

2. **Given Values**:
- Angle of dip, \( δ = 30^\circ \)
- Magnetic declination, \( α = 45^\circ \)

3. **Use the Formula for Apparent Dip**:
The apparent dip \( δ' \) can be calculated using the formula:
\[ \tan(δ') \cdot \cos(α) = \tan(δ) \]
Rearranging this gives:
\[ \tan(δ') = \frac{\tan(δ)}{\cos(α)} \]

4. **Calculate \( \tan(δ) \) and \( \cos(α) \)**:
- Calculate \( \tan(30^\circ) \):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \]
- Calculate \( \cos(45^\circ) \):
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707 \]

5. **Substitute the Values into the Formula**:
\[ \tan(δ') = \frac{\tan(30^\circ)}{\cos(45^\circ)} = \frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \]

6. **Calculate \( δ' \)**:
To find \( δ' \), take the arctangent:
\[ δ' = \tan^{-1}\left(\sqrt{\frac{2}{3}}\right) \]

7. **Final Result**:
The dip given by the dip circle is \( δ' \).