The horizontal component of earth's magnetic field is `sqrt3` times the vertical component . What is the value of angle of dip at this place ?

Answer

To find the angle of dip (also known as the magnetic inclination) when the horizontal component of Earth's magnetic field (BH) is \(\sqrt{3}\) times the vertical component (BV), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
We know that the horizontal component \( BH \) is given as:
\[ BH = \sqrt{3} \times BV \]

2. **Use the Definition of Angle of Dip**:
The angle of dip \( \delta \) can be defined using the tangent function:
\[ \tan(\delta) = \frac{BV}{BH} \]

3. **Substitute the Relationship**:
Substitute the expression for \( BH \) into the tangent equation:
\[ \tan(\delta) = \frac{BV}{\sqrt{3} \times BV} \]

4. **Simplify the Expression**:
The \( BV \) terms cancel out:
\[ \tan(\delta) = \frac{1}{\sqrt{3}} \]

5. **Find the Angle**:
To find the angle \( \delta \), we take the arctangent (inverse tangent) of both sides:
\[ \delta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

6. **Calculate the Angle**:
From trigonometric values, we know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Therefore:
\[ \delta = 30^\circ \]

### Final Answer:
The angle of dip at this place is \( 30^\circ \).