The true value of the angle of dip at a place | vedclass

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Question

(B) The relationship between the true dip $(\phi)$ and the apparent dip $(\phi')$ in a plane inclined at an angle $(\beta)$ to the magnetic meridian i

Vedclass · Accepted Answer

$	an^{-1}(2)$

Vedclass · Answer

(B) The relationship between the true dip $(\phi)$ and the apparent dip $(\phi')$ in a plane inclined at an angle $(\beta)$ to the magnetic meridian is given by the formula: $	an \phi' = \frac{	an \phi}{\cos \beta}$.Given: True dip $\phi = 60^o$ and inclination angle $\beta = 30^o$.Substituting the values: $	an \phi' = \frac{	an 60^o}{\cos 30^o}$.Since $	an 60^o = \sqrt{3}$ and $\cos 30^o = \frac{\sqrt{3}}{2}$,we get: $	an \phi' = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2$.Therefore,the apparent dip is $\phi' = 	an^{-1}(2)$.

The true value of the angle of dip at a place is $60^o$. The apparent dip in a plane inclined at an angle of $30^o$ with the magnetic meridian is:

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