First slide

Magnetic elements of Earth

Question

The plane of the dip circle is set in the geo-graphical meridian and the apparent dip is $θ_{1}$ . It is then set in a vertical plane perpendicular to the geo graphical meridian the apparent dip becomes $θ_{2}$ .The angle of declination $α$ at that place is given by

Moderate

A

$\tan α = \sqrt{\tan θ_{1} \tan θ_{2}}$
B

$α = \sqrt{\tan^{2} θ_{1} \tan^{2} θ_{2}}$
C

$\tan α = \frac{\tan θ_{1}}{\tan θ_{2}}$
D

$\tan α = \frac{\tan θ_{2}}{\tan θ_{1}}$

Solution

Let say $θ_{1} & θ_{2}$ are apparent dip in two different places.
$θ i s t r u e d i p a n d α i s t h e a n g l e o f d e c l i n a t i o n$
Case 1 : $\tan θ_{1} = \frac{\tan θ}{\cos α}$
Case 2 : $\tan θ_{2} = \frac{\tan θ}{\cos (90 - α)}$
$\begin{array}{l} A f t e r d i v i d i n g w e g e t, \frac{\tan θ_{1}}{\tan θ_{2}} = \frac{\sin α}{\cos α} = \tan α \\ \Rightarrow \frac{\tan θ_{1}}{\tan θ_{2}} = \tan α \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App