If θ _ 1 and θ _ 2 be the apparent angles of dip observed in two...

{\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 If θ _ 1 and θ _ 2 be the apparent angles of dip observed in two vertical planes at right angles to each other, then

Options

A $\cot^{2} \theta = \cot^{2} {\theta}_{1} + {\cot^{2} \theta}_{2}$
B $\tan^{2} \theta = {\tan^{2} \theta}_{1} + \tan^{2} {\theta}_{2}$
C $\cot^{2} \theta = \cot^{2} {\theta}_{1} - \cot^{2} {\theta}_{2}$
D $\tan^{2} \theta = \tan^{2} {\theta}_{1} - {\tan^{2} \theta}_{2}$

Correct answer

A. $\cot^{2} \theta = \cot^{2} {\theta}_{1} + {\cot^{2} \theta}_{2}$

Step-by-step solution

$∵$ apparent dip angle is given by

$tan {\theta}^{'} = \dfrac{tan \theta}{cos \alpha}$

where $\alpha$ is the angle between magnetic meridian and the apparent plane

If θ _ 1 and θ _ 2 be the apparent angles of dip observed in two vertical planes at right angles to each other, then

$∴ tan {\theta}_{1} = \dfrac{tan \theta}{cos \alpha}$ and $tan {\theta}_{2} = \dfrac{tan \theta}{\cos \left({90}^{o} - \alpha\right)} = \dfrac{tan \theta}{sin \alpha}$

$⇒ \sin^{2} \alpha + \cos^{2} \alpha = 1 ⇒ \cot^{2} {\theta}_{2} + \cot^{2} {\theta}_{1} = \cot^{2} \theta$

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