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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 \operatorname{dip} \theta is given by :-

Option: 1

\tan ^2 \theta=\tan ^2 \theta_1+\tan ^2 \theta_2


Option: 2

\cot ^2 \theta=\cot ^2 \theta_1-\cot ^2 \theta_2\cot ^2 \theta=\cot ^2 \theta_1-\cot ^2 \theta_2


Option: 3

\tan ^2 \theta=\tan ^2 \theta_1-\tan ^2 \theta_2^2


Option: 4

\cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2


Answers (1)

best_answer

If  \theta_1  and  \theta_2  are opparent angles of dip
Let \alpha  be the angle which one of the plane make with the magnetic meridian.
\tan \theta_1=\frac{\mathrm{V}}{\mathrm{H} \cos \alpha} \\

\text { i.e., } \cos \alpha=\frac{\mathrm{V}}{\mathrm{H} \tan \theta_1} \ldots \\.(i)

\tan \theta_2=\frac{\mathrm{V}}{\mathrm{H} \sin \alpha},

\text { i.e., } \sin \alpha=\frac{\mathrm{v}}{\mathrm{H} \tan \theta_2}........(ii)

Squaring and adding (i) and (ii), we get

\cos ^2 \alpha+\sin ^2 \alpha=\left(\frac{\mathrm{V}}{\mathrm{H}}\right)^2\left(\frac{1}{\tan ^2 \theta_1}+\frac{1}{\tan ^2 \theta_2}\right) \\
\text { i.e., } 1=\frac{\mathrm{V}^2}{\mathrm{H}^2}\left[\cot ^2 \theta_1+\cot ^2 \theta_2\right] \\
\text { or } \frac{\mathrm{H}^2}{\mathrm{~V}^2}=\cot ^2 \theta_1+\cot ^2 \theta_2 \\

\text { i.e., } \cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2.
 

Posted by

Ritika Jonwal

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