If the dip circle is set at 45^o to the magne | vedclass

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Question

$	an 	heta^{\prime}=\frac{	an 	heta}{\cos \alpha} \quad 	heta^{\prime} ightarrow$ apperent dip $=30$

$	an 30^{\circ}=\frac{	an 	heta}{\co

Vedclass · Accepted Answer

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

Vedclass · Answer

$	an 	heta^{\prime}=\frac{	an 	heta}{\cos \alpha} \quad 	heta^{\prime} ightarrow$ apperent dip $=30$

$	an 30^{\circ}=\frac{	an 	heta}{\cos 45^{\circ}} \quad 	heta ightarrow$ true dip $=?$

$	an 	heta=\frac{1}{\sqrt{3}} 	imes \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{6}} \quad \alpha=$ angle bet$^n$ two

planes $=45$

$	heta=	an ^{-1}\left(\frac{1}{\sqrt{6}}ight)$

If the dip circle is set at $45^o$ to the magnetic meridian, then the apparent dip is $30^o$. The true dip. of the place is

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