The values of apparent angles of dip in two places measured in two mutually perpendicular planes are `30^(@)` and `45^(@)`. Determine the true angle of dip at the place

Options

A. `tan^(-1)(4)`
B. `cot^(-1)(4)`
C. `tan^(-1)(2)`
D. `tan^(-1)(1/2)`

Answer

To determine the true angle of dip at a place where the apparent angles of dip in two mutually perpendicular planes are given as \(30^\circ\) and \(45^\circ\), we can follow these steps:

### Step 1: Understand the relationship between true dip and apparent dip
The true angle of dip (\(\delta\)) can be related to the apparent angles of dip (\(\delta_1\) and \(\delta_2\)) measured in two mutually perpendicular planes using the following relationships:
\[ \tan \delta = \frac{B_v}{B_h} \]
where \(B_v\) is the vertical component of the magnetic field and \(B_h\) is the horizontal component.

### Step 2: Set up equations for the apparent dips
For the first plane where the apparent dip is \(30^\circ\):
\[ \tan \delta_1 = \frac{B_v}{B_{h1}} = \tan 30^\circ \]
This gives:
\[ B_{h1} = B_v \cdot \sqrt{3} \]

For the second plane where the apparent dip is \(45^\circ\):
\[ \tan \delta_2 = \frac{B_v}{B_{h2}} = \tan 45^\circ \]
This gives:
\[ B_{h2} = B_v \]

### Step 3: Relate the horizontal components
Since the two planes are mutually perpendicular, we can express the horizontal component in terms of the angle \(\theta\):
\[ B_{h1} = B_h \cos \theta \]
\[ B_{h2} = B_h \sin \theta \]

### Step 4: Set up the equations
From the first plane:
\[ \tan 30^\circ = \frac{B_v}{B_h \cos \theta} \]
From the second plane:
\[ \tan 45^\circ = \frac{B_v}{B_h \sin \theta} \]

### Step 5: Divide the equations
Dividing the first equation by the second gives:
\[ \frac{\tan 30^\circ}{\tan 45^\circ} = \frac{B_h \sin \theta}{B_h \cos \theta} \]
This simplifies to:
\[ \tan 30^\circ = \tan \theta \]
Thus, we find that:
\[ \theta = 30^\circ \]

### Step 6: Substitute back to find true dip
Now, substituting \(\theta\) back into the equation for the true dip:
\[ \tan \delta = B_v \cdot \tan 30^\circ \cdot \cos 30^\circ \]
Using \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):
\[ \tan \delta = B_v \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{B_v}{2} \]

### Step 7: Find the angle
Thus, the true angle of dip is:
\[ \delta = \tan^{-1}\left(\frac{1}{2}\right) \]

### Final Answer
The true angle of dip at the place is:
\[ \delta \approx 26.57^\circ \]