A dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position, the angle of dip ia `40^(@)`. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of `30^(@)` with the magnetic meridian. In this position the needle will dip by an angle

Options

A. `40^(@)`
B. `30^(@)`
C. more than `40^(@)`
D. less than `40^(@)`

Explanation

To solve the problem, we need to find the new angle of dip (Δ) when the dip circle is rotated such that the plane in which the needle moves makes an angle of 30° with the magnetic meridian. We can use the relationship between the angles of dip and the angles of rotation.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial angle of dip (δ₁) = 40°
- Angle between the plane of the dip circle and the magnetic meridian (θ) = 30°

2. **Use the Cotangent Formula:**
The relationship between the angles can be expressed using cotangent:
\[ \cot(Δ) = \sqrt{\cot(δ₁) + \cot(θ)} \]

3. **Calculate Cotangent Values:**
- We need to find \(\cot(40°)\) and \(\cot(30°)\):
- \(\cot(40°) = \frac{1}{\tan(40°)}\)
- \(\cot(30°) = \sqrt{3}\)

4. **Substituting Values:**
- Calculate \(\cot(40°)\) using a calculator or trigonometric tables. Let's assume \(\cot(40°) \approx 1.19175\).
- Now substitute the values into the formula:
\[ \cot(Δ) = \sqrt{\cot(40°) + \cot(30°)} = \sqrt{1.19175 + \sqrt{3}} \]
- Since \(\sqrt{3} \approx 1.732\), we have:
\[ \cot(Δ) = \sqrt{1.19175 + 1.732} = \sqrt{2.92375} \]

5. **Calculate the Result:**
- Now, calculate \(\sqrt{2.92375}\):
\[ \cot(Δ) \approx 1.710 \]

6. **Find the Angle Δ:**
- To find Δ, take the cotangent inverse:
\[ Δ = \cot^{-1}(1.710) \]
- Using a calculator, we find:
\[ Δ \approx 25° \]

7. **Conclusion:**
- The new angle of dip (Δ) when the dip circle is rotated is approximately 25°, which is less than the initial angle of dip (40°).

### Final Answer:
The angle of dip when the dip circle is rotated is approximately **25°**.