Question

The angle of dip on a given meridian, making an angle ${30}^o$ with the magnetic meridian is equal to ${30}^o$. The true angle of dip at the given point is:

• A. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\tan }^{-1}\left(\sqrt{2}\right)$
• D. ${\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (A)

${\tan }^{-1}\left(\frac{1}{2}\right)$

$\text{Solution}$:

We know,

$tan\delta =\frac{V}{H}$

$\therefore$ by the given condition in the question,

$tan{30}^o=\frac{V}{Hcos{30}^o}$

Now, Divide (1) and (2), we get,

$\delta =ta{n}^{-1}\left(\frac{1}{2}\right)$