At a certain `B_(H) = (1)/(sqrt(3))` B dip, angle is:

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To solve the problem, we need to find the angle of dip (also known as magnetic inclination) given that \( B_H = \frac{1}{\sqrt{3}} B \). The angle of dip can be calculated using the relationship between the horizontal component of the magnetic field \( B_H \) and the vertical component \( B_V \).

### Step-by-Step Solution:

1. **Understand the Relationship**:
The angle of dip \( \theta \) is defined as:
\[ \tan(\theta) = \frac{B_V}{B_H} \]
where \( B_V \) is the vertical component and \( B_H \) is the horizontal component of the magnetic field.

2. **Identify the Given Values**:
From the problem, we have:
\[ B_H = \frac{1}{\sqrt{3}} B \]
We can express \( B_V \) in terms of \( B \):
\[ B_V = B \]

3. **Substitute the Values**:
Now substituting the values into the tangent formula:
\[ \tan(\theta) = \frac{B_V}{B_H} = \frac{B}{\frac{1}{\sqrt{3}} B} \]
Simplifying this gives:
\[ \tan(\theta) = \frac{B}{\frac{1}{\sqrt{3}} B} = \sqrt{3} \]

4. **Find the Angle**:
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}(\sqrt{3}) \]
We know that:
\[ \tan(60^\circ) = \sqrt{3} \]
Therefore:
\[ \theta = 60^\circ \]

5. **Conclusion**:
The angle of dip is:
\[ \theta = 60^\circ \]

### Final Answer:
The angle of dip is \( 60^\circ \).