Question

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

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

Answer 1

Answer: (A)

$\tan \theta^{'}=\frac{\tan \theta}{\cos \alpha}$
$\theta^{'}=\alpha=30^{\circ}$
$\therefore \tan \theta=(\tan 30)\left(\cos 30^{\circ}\right)=\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2}$