If `phi` is lattitude and `del` is dip at a place then

Options

A. `tan phi = (tan del)/(2)`
B. `tan del = (tan phi)/(2)`
C. `tan del = (1)/(tan phi)`
D. `tan ^(2) phi+ tan ^(2) del =1`

Answer

To solve the problem involving latitude (φ) and magnetic dip (δ), we will derive the relationship between these two quantities step by step.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- Latitude (φ) is the angle between the equatorial plane and a line from the center of the Earth to a point on its surface.
- Magnetic dip (δ) is the angle that the Earth's magnetic field makes with the horizontal plane.

2. **Using the Tangent Function**:
- We start with the relationship between latitude and magnetic dip. The tangent of the dip angle can be expressed in terms of latitude:
\[ \tan(δ) = k \tan(φ) \]
- Here, \(k\) is a constant that depends on the geographical location.

3. **Establishing the Relationship**:
- From the video transcript, we have the following relationships:
\[ \tan(φ) = \frac{1}{2} \tan(δ) \]
- This means that the tangent of latitude is half the tangent of the dip angle.

4. **Using the Sine and Cosine Relationships**:
- We can express the tangent functions in terms of sine and cosine:
\[ \tan(φ) = \frac{\sin(φ)}{\cos(φ)} \quad \text{and} \quad \tan(δ) = \frac{\sin(δ)}{\cos(δ)} \]

5. **Setting Up the Equations**:
- From the relationship established in step 3, we can substitute:
\[ \frac{\sin(φ)}{\cos(φ)} = \frac{1}{2} \cdot \frac{\sin(δ)}{\cos(δ)} \]
- Rearranging gives:
\[ 2 \sin(φ) \cos(δ) = \sin(δ) \cos(φ) \]

6. **Final Expression**:
- This leads us to the final relationship between latitude and dip:
\[ \tan(φ) = \frac{1}{2} \tan(δ) \]

### Conclusion:
The relationship between latitude (φ) and magnetic dip (δ) is given by:
\[ \tan(φ) = \frac{1}{2} \tan(δ) \]