If theta_1 and theta_2 are the apparent angle | vedclass

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Question

(A) Let $B_H$ and $B_V$ be the horizontal and vertical components of the Earth's magnetic field $\vec{B}$. The true angle of dip $	heta$ is given by

Vedclass · Accepted Answer

$\cot^2 	heta = \cot^2 	heta_1 + \cot^2 	heta_2$

Vedclass · Answer

(A) Let $B_H$ and $B_V$ be the horizontal and vertical components of the Earth's magnetic field $\vec{B}$. The true angle of dip $	heta$ is given by $	an 	heta = \frac{B_V}{B_H}$,or $\cot 	heta = \frac{B_H}{B_V}$.Suppose two mutually perpendicular vertical planes make angles $\alpha$ and $90^{\circ} - \alpha$ with the magnetic meridian. The vertical component $B_V$ remains the same in both planes,but the horizontal components in these planes are $B_H \cos \alpha$ and $B_H \sin \alpha$ respectively.The apparent angles of dip $	heta_1$ and $	heta_2$ in these planes are:$	an 	heta_1 = \frac{B_V}{B_H \cos \alpha} \implies \cot 	heta_1 = \frac{B_H \cos \alpha}{B_V}$$	an 	heta_2 = \frac{B_V}{B_H \sin \alpha} \implies \cot 	heta_2 = \frac{B_H \sin \alpha}{B_V}$Squaring and adding these equations:$\cot^2 	heta_1 + \cot^2 	heta_2 = \frac{B_H^2}{B_V^2} (\cos^2 \alpha + \sin^2 \alpha) = \frac{B_H^2}{B_V^2} = \cot^2 	heta$.Thus,$\cot^2 	heta = \cot^2 	heta_1 + \cot^2 	heta_2$.

If $\theta_1$ and $\theta_2$ are the apparent angles of dip observed in two vertical planes at right angles to each other,then the true angle of dip $\theta$ is given by:

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