If $ heta_1$ and $ heta_2$ be the apparent angles of dip observed in two vertical planes at right angles to

Question

If $	heta_1$ and $	heta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $	heta$ is given by:

Accepted Answer

1. **True Dip ($	heta$):** The true angle of dip is related to the vertical component ($V$) and horizontal component ($H$) of Earth's magnetic field in the magnetic meridian by $	an 	heta = \frac{V}{H}$ or $\cot 	heta = \frac{H}{V}$.
2. **Apparent Dip:** In a vertical plane making an angle $\alpha$ with the magnetic meridian, the effective horizontal component is $H' = H \cos \alpha$, while the vertical component $V$ remains the same. The apparent dip $	heta_1$ is given by:
   $$ 	an 	heta_1 = \frac{V}{H \cos \alpha} \Rightarrow \cot 	heta_1 = \frac{H \cos \alpha}{V} = \cot 	heta \cos \alpha $$
   $$ \cos \alpha = \frac{\cot 	heta_1}{\cot 	heta} $$
3. **Second Plane:** Since the second plane is perpendicular to the first, it makes an angle $(90^\circ - \alpha)$ with the meridian. The apparent dip $	heta_2$ is:
   $$ 	an 	heta_2 = \frac{V}{H \cos(90^\circ - \alpha)} = \frac{V}{H \sin \alpha} \Rightarrow \cot 	heta_2 = \frac{H \sin \alpha}{V} = \cot 	heta \sin \alpha $$
   $$ \sin \alpha = \frac{\cot 	heta_2}{\cot 	heta} $$
4. **Derivation:** Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:
   $$ \left( \frac{\cot 	heta_2}{\cot 	heta} ight)^2 + \left( \frac{\cot 	heta_1}{\cot 	heta} ight)^2 = 1 $$
   $$ \frac{\cot^2 	heta_2 + \cot^2 	heta_1}{\cot^2 	heta} = 1 $$
   $$ \cot^2 	heta = \cot^2 	heta_1 + \cot^2 	heta_2 $$

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