The real angle of dip, if a magnet is suspended at an angle of `30^(@)` to the magnetic meridian and the dip needle makes an angle of `45^(@)` with horizontal, is:

Options

A. `tan^(-1)(3//sqrt2)`
B. `tan^(-1)(sqrt3)`
C. `tan^(-1)(3//sqrt2)`
D. `tan^(-1)(2//sqrt3)`

Explanation

To find the real angle of dip (Δ') when a magnet is suspended at an angle of 30° to the magnetic meridian and the dip needle makes an angle of 45° with the horizontal, we can follow these steps:

### Step 1: Understand the relationship between the angles
The relationship between the angle of dip (Δ), the angle at which the magnet is suspended (θ), and the real angle of dip (Δ') can be expressed using the formula:
\[ \tan(\Delta') = \frac{\tan(\Delta)}{\cos(\theta)} \]
where:
- Δ = angle of dip (45° in this case)
- θ = angle made by the magnet with the magnetic meridian (30° in this case)

### Step 2: Substitute the known values
We know that:
- Δ = 45°
- θ = 30°

Thus, we can substitute these values into the formula:
\[ \tan(\Delta') = \frac{\tan(45°)}{\cos(30°)} \]

### Step 3: Calculate the values of tan(Δ) and cos(θ)
We know:
- \(\tan(45°) = 1\)
- \(\cos(30°) = \frac{\sqrt{3}}{2}\)

### Step 4: Substitute these values into the equation
Now substituting these values into the equation gives us:
\[ \tan(\Delta') = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]

### Step 5: Find the angle Δ'
To find the angle Δ', we take the arctangent (inverse tangent) of both sides:
\[ \Delta' = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \]

### Step 6: Conclusion
Thus, the real angle of dip (Δ') is:
\[ \Delta' = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \]

### Final Answer
The real angle of dip is \( \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \).

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