Question

A magnetic needle freed to rotate around a fixed vertical plane stay at an angle ${60}^0$ with the horizontal . If the dip at that place is ${37}^0$, find the angle of the fixed vertical plane with the meridian.

• A. $\theta ={\cos }^{-1}\frac{\sqrt{3}}{5}$
• B. $\theta ={\cos }^{-1}\frac{1}{\sqrt{3}}$
• C. $\theta ={\cos }^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• D. $\theta ={\cos }^{-1}\frac{2}{3}$

Answer 1

Answer: (C)

$\theta ={\cos }^{-1}\left(\frac{\sqrt{3}}{4}\right)$

$\tan {\delta }_1=\frac{B_V}{B_H\cos \theta }=\frac{\tan \delta }{\cos \theta }$ or
$\cos \theta =\frac{\tan \delta }{\tan {\delta }_1}=\frac{\tan 37}{\tan 60}=\frac{3/4}{\sqrt{3}}=\frac{\sqrt{3}}{4}$