If theta _1 and theta_2 be the apparent angle | vedclass

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Question

Let $B_H$ and $B_v$ be the horizontal and vertical components of earth's magnetic field $\vec B$. Since $	heta$ is the angle of dip

$	an 	he

Vedclass · Accepted Answer

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

Vedclass · Answer

Let $B_H$ and $B_v$ be the horizontal and vertical components of earth's magnetic field $\vec B$. Since $	heta$ is the angle of dip

$	an 	heta = \frac{B_V}{B_H}$

 or $cot	heta = \frac{B_H}{B_V}$

Suppose planes $1$ and $2$ are two mutually perpendicular planes and respectively make angles $	heta$ and $90&deg;- 	heta$ with the magnetic meridian. The vertical components of earth's magnetic field remain same in the two planes but the effective horizontal components in the planes will be

$B_{1}=B_{H} \cos 	heta 	ext { and } B_{2}=B_{H} \sin 	heta$

The angles of $\operatorname{dip} 	heta_{1}$ and $	heta_{2}$ in the two planes are given by

$	an 	heta_{1}=\frac{B_{V}}{B_{1}}$

$	an 	heta_{1}=\frac{B_{V}}{B_{H} \cos 	heta}$

or $\quad \cot 	heta_{1}=\frac{B_{H} \sin 	heta}{B_{V}}.........(ii)$

$	ext { Similarly, } \cot 	heta_{2}=\frac{B_{H} \sin 	heta}{B_{V}}.........(iii)$

From eqns. $(ii)$ and $(iii)$

$\cot ^{2} 	heta_{1}+\cot ^{2} 	heta_{2}=\frac{B_{H}^{2}}{B_{V}^{2}}\left(\cos ^{2} 	heta+\sin ^{2} 	hetaight)=\frac{B_{H}^{2}}{B_{V}^{2}}$

$	herefore \quad \cot ^{2} 	heta_{1}+\cot ^{2} 	heta_{2}=\cot ^{2} 	heta \quad[	ext { from eqn. (i) }]$

If $\theta _1$ and $\theta_2$ be 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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