Question

At ${45}^o$ to the magnetic meridian , apparent dip is ${30}^o$. Find true dip :

• A. ${\tan }^{-1}\frac{1}{\sqrt{2}}$
• B. ${\tan }^{-1}\sqrt{6}$
• C. ${\tan }^{-1}\sqrt{2}$
• D. ${\tan }^{-1}\frac{1}{\sqrt{6}}$

Answer 1

The correct option is D ${\tan }^{-1}\frac{1}{\sqrt{6}}$

$\tan \delta =\tan {\delta }^{\prime }\cos \theta =\tan {30}^o\cos {45}^o$
(standard equation formed by the information given in the question)
$=\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{6}}$