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(C) Let $\delta$ be the true angle of dip and $	heta$ be the magnetic declination. When the dip circle is in the magnetic meridian,the apparent dip i

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$	heta=\operatorname{Tan}^{-1}\left(\frac{	an \delta_1}{	an \delta_2}ight)$

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(C) Let $\delta$ be the true angle of dip and $	heta$ be the magnetic declination. When the dip circle is in the magnetic meridian,the apparent dip is $\delta$. However,the problem states the plane is set in the geographic meridian. Let $	heta$ be the angle between the geographic meridian and the magnetic meridian. The apparent dip $\delta_1$ in a plane making an angle $	heta$ with the magnetic meridian is given by $	an \delta_1 = \frac{	an \delta}{\cos 	heta}$. The apparent dip $\delta_2$ in a plane perpendicular to the first (making an angle $90^\circ - 	heta$ with the magnetic meridian) is given by $	an \delta_2 = \frac{	an \delta}{\cos(90^\circ - 	heta)} = \frac{	an \delta}{\sin 	heta}$. From these two equations,we have $	an \delta = 	an \delta_1 \cos 	heta$ and $	an \delta = 	an \delta_2 \sin 	heta$. Equating them: $	an \delta_1 \cos 	heta = 	an \delta_2 \sin 	heta$. Therefore,$\frac{\sin 	heta}{\cos 	heta} = \frac{	an \delta_1}{	an \delta_2}$,which implies $	an 	heta = \frac{	an \delta_1}{	an \delta_2}$. Thus,$	heta = \operatorname{Tan}^{-1}\left(\frac{	an \delta_1}{	an \delta_2}ight)$.

The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_1$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_2$. The declination $\theta$ at the place is

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