The original value of the angle of dip at a place is `45^(@)` what will be the apparent dip angle on a plane inclined at `60^(@)` with the magnetic meridian ?

Options

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

Answer

To solve the problem of finding the apparent dip angle when the original angle of dip is \(45^\circ\) and the plane is inclined at \(60^\circ\) with the magnetic meridian, we can follow these steps:

### Step 1: Understand the relationship between true dip and apparent dip
The true dip angle (\(\delta\)) is the angle that the magnetic field makes with the horizontal. The apparent dip angle (\(\delta'\)) is affected by the inclination of the plane with respect to the magnetic meridian. The relationship can be expressed as:
\[ \tan \delta' = \frac{\tan \delta}{\cos \theta} \]
where \(\theta\) is the angle of inclination of the plane with the magnetic meridian.

### Step 2: Identify the values
From the problem:
- The true dip angle \(\delta = 45^\circ\)
- The angle of inclination \(\theta = 60^\circ\)

### Step 3: Calculate \(\tan \delta\)
Using the known value:
\[ \tan 45^\circ = 1 \]

### Step 4: Calculate \(\cos \theta\)
Now, calculate \(\cos 60^\circ\):
\[ \cos 60^\circ = \frac{1}{2} \]

### Step 5: Substitute the values into the formula
Now substitute \(\tan \delta\) and \(\cos \theta\) into the formula for \(\tan \delta'\):
\[ \tan \delta' = \frac{\tan 45^\circ}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2 \]

### Step 6: Find the apparent dip angle
Now, to find \(\delta'\), we take the arctangent:
\[ \delta' = \tan^{-1}(2) \]

### Final Answer
Thus, the apparent dip angle is:
\[ \delta' = \tan^{-1}(2) \]