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Question

(B) Let $\delta$ be the true dip angle and $\delta'$ be the apparent dip angle in a plane making an angle $	heta = 60^{\circ}$ with the magnetic meri

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Let $\delta$ be the true dip angle and $\delta'$ be the apparent dip angle in a plane making an angle $	heta = 60^{\circ}$ with the magnetic meridian.The relationship between the true dip and apparent dip is given by $	an \delta' = \frac{	an \delta}{\cos 	heta}$.Given $\delta' = 	an^{-1}\left(\frac{2}{\sqrt{3}}ight)$,so $	an \delta' = \frac{2}{\sqrt{3}}$.Substituting the values: $\frac{2}{\sqrt{3}} = \frac{	an \delta}{\cos 60^{\circ}}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $\frac{2}{\sqrt{3}} = \frac{	an \delta}{1/2} = 2 	an \delta$.Therefore,$	an \delta = \frac{1}{\sqrt{3}}$.This implies $\delta = 30^{\circ}$.

$A$ magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{\circ}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$ with the horizontal,the true dip value at that place is: (in $^{\circ}$)

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