The dip angle in a vertical plane at an angle | vedclass

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Question

(D) Let $	heta$ be the actual dip angle and $	heta'$ be the apparent dip angle in a vertical plane making an angle $\alpha$ with the magnetic meridi

Vedclass · Accepted Answer

$	an^{-1} \left( \sqrt{\frac{3}{2}} ight)$

Vedclass · Answer

(D) Let $	heta$ be the actual dip angle and $	heta'$ be the apparent dip angle in a vertical plane making an angle $\alpha$ with the magnetic meridian.The relationship between the apparent dip and the actual dip is given by the formula: $	an 	heta' = \frac{	an 	heta}{\cos \alpha}$.Given: $\alpha = \cos^{-1} \left( \frac{1}{\sqrt{2}} ight)$,so $\cos \alpha = \frac{1}{\sqrt{2}}$.Given: $	heta' = 60^\circ$,so $	an 	heta' = 	an 60^\circ = \sqrt{3}$.Substituting these values into the formula:$\sqrt{3} = \frac{	an 	heta}{1/\sqrt{2}}$$	an 	heta = \sqrt{3} 	imes \frac{1}{\sqrt{2}} = \sqrt{\frac{3}{2}}$.Therefore,the actual dip angle $	heta = 	an^{-1} \left( \sqrt{\frac{3}{2}} ight)$.

The dip angle in a vertical plane at an angle of $\cos^{-1} \left( \frac{1}{\sqrt{2}} \right)$ from the magnetic meridian is $60^\circ$. Find the actual dip angle at that place.

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