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Q.

If $ϕ_{1} and ϕ_{2}$ be the angles of dip observed in two vertical planes at right angles to each other and $ϕ$ be the true angle of dip, then

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a

$\cot^{2} ϕ = \cot^{2} ϕ_{1} + \cot^{2} ϕ_{2}$

b

$\sec^{2} ϕ = \sec^{2} ϕ_{1} + \sec^{2} ϕ_{2}$

c

$\cos^{2} ϕ = \cos^{2} ϕ_{1} + \cos^{2} ϕ_{2}$

d

$\tan^{2} ϕ = \tan^{2} ϕ_{1} + \tan^{2} ϕ_{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} \tan ϕ_{1} = \frac{\tan ϕ}{\cos θ}; \cos θ = \frac{\tan ϕ}{\tan ϕ_{1}} \\ \cos (90 - θ) = \sin θ = \frac{\tan ϕ}{\tan ϕ_{2}} \\ \sin^{2} θ + \cos^{2} θ = 1 = \tan^{2} ϕ [\cot^{2} ϕ_{1} + \cot^{2} ϕ_{2}] \\ \cot^{2} ϕ = \cot^{2} ϕ_{1} + \cot^{2} ϕ_{2} \end{array}$

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If ϕ1 and ϕ2 be the angles of dip observed in two vertical planes at right angles to each other and ϕ be the true angle of dip, then