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(C) Let $\phi$ be the true dip angle and $	heta$ be the magnetic declination. When the dip circle is in the geographic meridian,the angle between the

Vedclass · Accepted Answer

$	an ^{-1}(\frac{	an \delta_1}{	an \delta_2})$

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(C) Let $\phi$ be the true dip angle and $	heta$ be the magnetic declination. When the dip circle is in the geographic meridian,the angle between the plane and the magnetic meridian is $	heta$. The apparent dip $\delta_1$ is given by $	an \delta_1 = \frac{	an \phi}{\cos 	heta}$.When the plane is perpendicular to the geographic meridian,the angle between the plane and the magnetic meridian is $(90^\circ - 	heta)$. The apparent dip $\delta_2$ is given by $	an \delta_2 = \frac{	an \phi}{\cos(90^\circ - 	heta)} = \frac{	an \phi}{\sin 	heta}$.From these two equations,we have $	an \phi = 	an \delta_1 \cos 	heta$ and $	an \phi = 	an \delta_2 \sin 	heta$.Equating the two expressions for $	an \phi$: $	an \delta_1 \cos 	heta = 	an \delta_2 \sin 	heta$.Rearranging gives $\frac{\sin 	heta}{\cos 	heta} = \frac{	an \delta_1}{	an \delta_2}$,which implies $	an 	heta = \frac{	an \delta_1}{	an \delta_2}$.Therefore,$	heta = 	an ^{-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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