Apparent dips when the dip circle is placed into two mutually perpendicular directions are ${30^0}$ and ${45^0}$. What is the actual dip at that place?
A. ${\tan ^{ - 1}}2$
B. ${\tan ^{ - 1}}(\dfrac{1}{2})$
C. ${\tan ^{ - 1}}\sqrt 2 $
D. ${60^ \circ }$

Hint: Dip circles are also known as dip needles. These are the devices which are used to measure the angle between the horizon and the earth’s magnetic field. This angle is called the dip angle at a particular place.

Complete step by step answer:
It's given that in the question that there are two dip circles which are perpendicular to each other and their own respective directions are ${30^0}$ and ${45^0}$. Let us assume that the angle made by the first dip circle is denoted by ${\theta _1}$ and the angle made by the second dip circle is denoted by ${\theta _2}$ .

The magnitude of ${\theta _1}$ is ${\theta _1} = {30^0}$ and
The magnitude of ${\theta _2}$ is ${\theta _2} = {45^0}$.
Let us find the value of their cotangent which is represented as
$\cot {30^0} = \sqrt 3 $ And
$\Rightarrow \cot {45^0} = 1$
Since, both the angle is perpendicular to each other hence Using Pythagoras theorem
We can write as:
${\cot ^2}\theta = {\cot ^2}{\theta _1} + {\cot ^2}{\theta _2}$
Where $\theta $ is the net angle made at that place
So,
${\cot ^2}\theta = {\cot ^2}{30^0} + {\cot ^2}{45^0}$
$\Rightarrow {\cot ^2}\theta = 3 + 1$
$\Rightarrow \cot \theta = 2$
Since cotangent is the inverse of tangent trigonometric function so
$\tan \theta = \dfrac{1}{2}$
Taking the inverse of tangent function we will get,
$\therefore \theta = {\tan ^{ - 1}}(\dfrac{1}{2})$

Hence, the correct option is B.

Note:Dip circles were usually used in the working of mining, surveying and as well as to study magnetism. It gives an accurate angle which is formed between the earth’s horizon and the earth’s magnetic field at a particular point and mostly it’s calculated in trigonometric tangent function which is $\tan \theta $.