Question

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

• A. ${\cot }^2\varphi ={\cot }^2{\varphi }_1+{\cot }^2{\varphi }_2$
• B. ${\tan }^2\varphi ={\tan }^2{\varphi }_1+{\tan }^2{\varphi }_2$
• C. ${\tan }^2\varphi ={\tan }^2{\varphi }_1-{\tan }^2{\varphi }_2$
• D. ${\cot }^2\varphi ={\cot }^2{\varphi }_1-{\cot }^2{\varphi }_2$

Answer 1

The correct option is A ${\cot }^2\varphi ={\cot }^2{\varphi }_1+{\cot }^2{\varphi }_2$


Let $\alpha$ be the angle which one of the planes make with the magnetic meridian the other plane makes an angle $(90°-\alpha )$ with it. The components of $H$ in these planes will be $H\cos \alpha$   and $H\sin \alpha$   respectively. If ${\varphi }_1$ and ${\varphi }_2$ are the apparent dips in these two planes, then

$\tan {\varphi }_1=\frac{V}{H\cos \alpha }$ i.e.

$\cos \alpha =\frac{V}{H\tan {\varphi }_1}$ ... (i)

$\tan {\varphi }_2=\frac{V}{H\sin \alpha }$ i.e.

$\sin \alpha =\frac{V}{H\tan {\varphi }_2}$ ... (ii)

Squaring and adding (i) and (ii), we get

${\cos }^2\alpha +{\sin }^2\alpha ={\left(\frac{V}{H}\right)}^2(\frac{1}{{\tan }^2{\varphi }_1}+\frac{1}{{\tan }^2{\varphi }_2})$

i.e. $1=\frac{V^2}{H^2}({\cot }^2{\varphi }_1+{\cot }^2{\varphi }_2)$

or $\frac{H^2}{V^2}={\cot }^2{\varphi }_1+{\cot }^2{\varphi }_2$ i.e.,

${\cot }^2\varphi ={\cot }^2{\varphi }_1+{\cot }^2\varphi$

This is the required result.
Hence, option $\left(D\right)$ is correct.