Question

If ${\delta }_1$ and ${\delta }_2$ be the angles of dip observed in two vertical planes at right angles to each other and $\delta$ be the true angle of dip, then

• A. ${\cos }^2\delta ={\cos }^2{\delta }_1+{\cos }^2{\delta }_2$
• B. ${\sec }^2\delta ={\sec }^2{\delta }_1+{\sec }^2{\delta }_2$
• C. ${\tan }^2\delta ={\tan }^2{\delta }_1+{\tan }^2{\delta }_2$
• D. ${\cot }^2\delta ={\cot }^2{\delta }_1+{\cot }^2{\delta }_2$

Answer 1

The correct option is D ${\cot }^2\delta ={\cot }^2{\delta }_1+{\cot }^2{\delta }_2$

Let $\alpha$ be the angle made by one of the planes with magnetic meridian, then the other plane makes an angle $(90-\alpha )$. The components of $B$ in these planes will be $B_H\cos \alpha$ and $B_H\sin \alpha$, as shown in diagram.


Given, ${\delta }_1$ and ${\delta }_2$ are the apparent dips in these planes then,

$\tan {\delta }_1=\frac{B_V}{(B_H{)}_1}=\frac{B_V}{B_H\cos \alpha }$

$\Rightarrow \cos \alpha =\frac{B_V}{\left(\tan {\delta }_1\right)B_H}$

$\Rightarrow \cos \alpha =\frac{B_V}{B_H}\cot {\delta }_1.....\left(i\right)$

$\tan {\delta }_2=\frac{B_V}{(B_H{)}_2}=\frac{B_V}{B_H\sin \alpha }$

$\Rightarrow \sin \alpha =\frac{B_V}{B_H\left(\tan {\delta }_2\right)}$

$\Rightarrow \sin \alpha =\frac{B_V}{B_H}\cot {\delta }_2.....\left(ii\right)$

Squaring and adding equations $\left(i\right)$ and $\left(ii\right)$

${\cos }^2\alpha +{\sin }^2\alpha ={\left(\frac{B_V}{B_H}\right)}^2{\cot }^2{\delta }_1+{\left(\frac{B_V}{B_H}\right)}^2{\cot }^2{\delta }_2$

$\therefore 1={\left(\frac{B_V}{B_H}\right)}^2({\cot }^2{\delta }_1+{\cot }^2{\delta }_2)$

${\cot }^2{\delta }_1+{\cot }^2{\delta }_2=\frac{1}{{\tan }^2\delta }$

$\therefore {\cot }^2{\delta }_1+{\cot }^2{\delta }_2={\cot }^2\delta (\because \frac{B_V}{B_H}=\tan \delta )$

Hence, option $\left(D\right)$ is the correct answer.

Why this Question? Note: This is a very standard result from the NEET point of view, and students are expected to remember the result.