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

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Question

$A 	an \delta=	an \delta^{\prime} \cos 	heta$

$=	an 45^{\circ} \cos 30^{\circ}$

$	an \delta=1 	imes \frac{\sqrt{3}}{2}$

$\delta=	an ^{

Vedclass · Accepted Answer

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

Vedclass · Answer

$A 	an \delta=	an \delta^{\prime} \cos 	heta$

$=	an 45^{\circ} \cos 30^{\circ}$

$	an \delta=1 	imes \frac{\sqrt{3}}{2}$

$\delta=	an ^{-1}\left(\frac{\sqrt{3}}{2}ight)$

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

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