Question

A magnet hung at $45^{\circ}$ with magnetic meridian makes an angle of $60^{\circ}$ with the horizontal. The actual value of the angle of dip is

• A. $\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)$
• B. $\tan ^{-1}(\sqrt{6})$
• C. $\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)$
• D. $\tan ^{-1}\left(\sqrt{\frac{1}{2}}\right)$

Answer 1

Answer: (A)

$ \tan \theta^{\prime}=\frac{\tan \theta}{\cos \alpha} $
$\theta^{\prime}=60^{\circ} $
$ \alpha=45^{\circ} $
$ \sqrt{3}=\frac{\tan \theta}{\frac{1}{\sqrt{2}}} $
$ \tan \theta=\sqrt{\frac{3}{2}} $
$ \theta=\tan ^{-1} \sqrt{\frac{3}{2}}$