If {phi_1} and {phi_2} are the angles of dip | vedclass

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(D) Let $\alpha$ be the angle which one of the planes makes with the magnetic meridian. The other plane makes an angle $(90^\circ - \alpha)$ with it.T

Vedclass · Accepted Answer

${\cot ^2}\phi= {\cot ^2}{\phi_1} + {\cot ^2}{\phi_2}$

Vedclass · Answer

(D) Let $\alpha$ be the angle which one of the planes makes with the magnetic meridian. The other plane makes an angle $(90^\circ - \alpha)$ with it.The horizontal components of the Earth's magnetic field $H$ in these two planes are $H_1 = H \cos \alpha$ and $H_2 = H \sin \alpha$ respectively.If $\phi_1$ and $\phi_2$ are the apparent dips in these two planes,and $V$ is the vertical component of the Earth's magnetic field,then:$	an \phi_1 = \frac{V}{H \cos \alpha} \implies \cos \alpha = \frac{V}{H 	an \phi_1}$ ..... $(i)$$	an \phi_2 = \frac{V}{H \sin \alpha} \implies \sin \alpha = \frac{V}{H 	an \phi_2}$ ..... $(ii)$Squaring and adding $(i)$ and $(ii)$,we get:$\cos^2 \alpha + \sin^2 \alpha = \left( \frac{V}{H} ight)^2 \left( \frac{1}{	an^2 \phi_1} + \frac{1}{	an^2 \phi_2} ight)$$1 = \frac{V^2}{H^2} (\cot^2 \phi_1 + \cot^2 \phi_2)$Since the true dip $\phi$ satisfies $	an \phi = \frac{V}{H}$,we have $\cot^2 \phi = \frac{H^2}{V^2}$.Substituting this into the equation,we get:$\cot^2 \phi = \cot^2 \phi_1 + \cot^2 \phi_2$.

If ${\phi_1}$ and ${\phi_2}$ are the angles of dip observed in two vertical planes at right angles to each other and ${\phi}$ is the true angle of dip,then:

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