If the dip circle is set at 45^{circ} to the | vedclass

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(B) The relation between the apparent dip $(	heta^{\prime})$ and the true dip $(	heta)$ when the dip circle is at an angle $\alpha$ to the magnetic

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{1}{\sqrt{6}}ight)$

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(B) The relation between the apparent dip $(	heta^{\prime})$ and the true dip $(	heta)$ when the dip circle is at an angle $\alpha$ to the magnetic meridian is given by: $	an 	heta^{\prime} = \frac{	an 	heta}{\cos \alpha}$ Given: Apparent dip $	heta^{\prime} = 30^{\circ}$ Angle with magnetic meridian $\alpha = 45^{\circ}$ Substituting these values into the formula: $	an 30^{\circ} = \frac{	an 	heta}{\cos 45^{\circ}}$ $\frac{1}{\sqrt{3}} = \frac{	an 	heta}{1/\sqrt{2}}$ $	an 	heta = \frac{1}{\sqrt{3}} 	imes \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{6}}$ Therefore,the true dip $	heta = 	an^{-1}\left(\frac{1}{\sqrt{6}}ight)$.

If the dip circle is set at $45^{\circ}$ to the magnetic meridian,then the apparent dip is $30^{\circ}$. The true dip of the place is:

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