If `theta_1` and `theta_2` be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip `theta` is given by

Options

A. `cot^(2)theta=cot^(2)theta_(1)+cot^(2)theta_(2)`
B. `tan^(2)theta=tan^(2)theta_(1)+tan^(2)theta_(2)`
C. `cot^(2)theta=cot^(2)theta_(1)-cot^(2)theta_(2)`
D. `tan^(2)theta=tan^(2)theta_(1)-tan^(2)theta_(2)`

Answer

To find the true angle of dip \( \theta \) given the apparent angles of dip \( \theta_1 \) and \( \theta_2 \) observed in two vertical planes at right angles to each other, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Angles of Dip**:
- The angles \( \theta_1 \) and \( \theta_2 \) are the apparent angles of dip in two vertical planes that are perpendicular to each other.

2. **Relating the Angles to Components of the Magnetic Field**:
- The tangent of the angle of dip can be expressed in terms of the vertical and horizontal components of the Earth's magnetic field:
\[ \tan \theta_1 = \frac{V_1}{H_1} \]
\[ \tan \theta_2 = \frac{V_1}{H_2} \]
- Here, \( V_1 \) is the vertical component of the magnetic field, and \( H_1 \) and \( H_2 \) are the horizontal components in the respective planes.

3. **Using the Right Triangle Relationship**:
- Since the two planes are at right angles, we can use the Pythagorean theorem to relate the horizontal components:
\[ H^2 = H_1^2 + H_2^2 \]

4. **Expressing Horizontal Components in Terms of Tangents**:
- From the definitions of tangent, we can express \( H_1 \) and \( H_2 \) in terms of \( V_1 \):
\[ H_1 = \frac{V_1}{\tan \theta_1} \]
\[ H_2 = \frac{V_1}{\tan \theta_2} \]

5. **Substituting into the Pythagorean Theorem**:
- Substitute \( H_1 \) and \( H_2 \) into the equation for \( H \):
\[ H^2 = \left(\frac{V_1}{\tan \theta_1}\right)^2 + \left(\frac{V_1}{\tan \theta_2}\right)^2 \]

6. **Expressing the True Angle of Dip**:
- The true angle of dip \( \theta \) relates to the total horizontal component \( H \) and the vertical component \( V_1 \):
\[ H = \frac{V_1}{\tan \theta} \]
- Therefore, we can write:
\[ \left(\frac{V_1}{\tan \theta}\right)^2 = \left(\frac{V_1}{\tan \theta_1}\right)^2 + \left(\frac{V_1}{\tan \theta_2}\right)^2 \]

7. **Cancelling \( V_1^2 \)**:
- Cancel \( V_1^2 \) from both sides:
\[ \frac{1}{\tan^2 \theta} = \frac{1}{\tan^2 \theta_1} + \frac{1}{\tan^2 \theta_2} \]

8. **Using Cotangent**:
- Recall that \( \frac{1}{\tan} = \cot \):
\[ \cot^2 \theta = \cot^2 \theta_1 + \cot^2 \theta_2 \]

9. **Final Expression**:
- The true angle of dip \( \theta \) is given by:
\[ \cot^2 \theta = \cot^2 \theta_1 + \cot^2 \theta_2 \]