The true value of angle of dip at a place is `60^(@)` . The apparent angle of dip , when a magnetic needle is rotated through `30^(@)` from the magnetic meridian at the same place , is

Options

A. `tan^(-1) (2)`
B. `tan ^(-1) ((1)/(2))`
C. `tan^(-1) ((3)/(2))`
D. `tan^(-1) ((4)/(3))`

Answer

To solve the problem, we need to find the apparent angle of dip when a magnetic needle is rotated through 30 degrees from the magnetic meridian, given that the true angle of dip is 60 degrees.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The true angle of dip (δ) is given as 60 degrees.
- The apparent angle of dip (δ') needs to be calculated when the needle is rotated through an angle θ = 30 degrees.

2. **Use the Formula for Angle of Dip**:
- The angle of dip is related to the vertical (Bv) and horizontal (Bh) components of the Earth's magnetic field.
- We know that:
\[ \tan(δ) = \frac{Bv}{Bh} \]

3. **Calculate the Apparent Angle of Dip**:
- When the needle is rotated, the vertical component (Bv) remains unchanged, but the horizontal component changes to Bh'.
- The new horizontal component can be expressed as:
\[ Bh' = Bh \cos(θ) \]
- Therefore, the apparent angle of dip can be expressed as:
\[ \tan(δ') = \frac{Bv}{Bh'} = \frac{Bv}{Bh \cos(θ)} \]

4. **Relate the Two Angles**:
- From the previous equations, we can set up the relationship:
\[ \tan(δ') = \frac{Bv}{Bh \cos(θ)} = \frac{Bv}{Bh} \cdot \frac{1}{\cos(θ)} = \tan(δ) \cdot \frac{1}{\cos(θ)} \]

5. **Substitute Known Values**:
- Substitute δ = 60 degrees and θ = 30 degrees into the equation:
\[ \tan(δ') = \tan(60^\circ) \cdot \frac{1}{\cos(30^\circ)} \]
- We know:
- \(\tan(60^\circ) = \sqrt{3}\)
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)

6. **Calculate the Result**:
- Substitute these values into the equation:
\[ \tan(δ') = \sqrt{3} \cdot \frac{1}{\frac{\sqrt{3}}{2}} = \sqrt{3} \cdot \frac{2}{\sqrt{3}} = 2 \]
- Therefore, the apparent angle of dip is:
\[ δ' = \tan^{-1}(2) \]

### Final Answer:
The apparent angle of dip when the magnetic needle is rotated through 30 degrees from the magnetic meridian is \( \tan^{-1}(2) \).

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