At an angle of 30^{circ} to the magnetic meri | vedclass

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Question

(B) The relationship between the true dip $(\delta)$ and the apparent dip $(\delta^{\prime})$ at an angle $	heta$ to the magnetic meridian is given b

Vedclass · Accepted Answer

$	an^{-1} \frac{\sqrt{3}}{2}$

Vedclass · Answer

(B) The relationship between the true dip $(\delta)$ and the apparent dip $(\delta^{\prime})$ at an angle $	heta$ to the magnetic meridian is given by the formula: $	an \delta^{\prime} = \frac{	an \delta}{\cos 	heta}$.Rearranging for the true dip: $	an \delta = 	an \delta^{\prime} \cos 	heta$.Given: $\delta^{\prime} = 45^{\circ}$ and $	heta = 30^{\circ}$.Substituting the values: $	an \delta = 	an 45^{\circ} \cos 30^{\circ}$.Since $	an 45^{\circ} = 1$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we get: $	an \delta = 1 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.Therefore,the true dip is $\delta = 	an^{-1} \left( \frac{\sqrt{3}}{2} ight)$.

At an angle of $30^{\circ}$ to the magnetic meridian,the apparent dip is $45^{\circ}$. Find the true dip.

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