If θ 1 and θ 2 be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by (a) tan2 θ = tan2 θ 1 + tan2 θ 2 (b) cot2 θ = cot2 θ 1 - cot2 θ 2 (c) tan2 θ = tan2 θ 1 - tan2 θ 2 (d) cot2 θ = cot2 θ 1 + cot2 θ 2

Answer: (d). In perpendicular planes: tan θ₁ = (V)/(Hcos α), tan θ₂ = (V)/(Hsin α). cot ²θ₁ + cot ²θ₂ = (H²(cos ²α + sin ²α))/(V²) = (H²)/(V²) = cot ²θ So the answer is (d) cot ²θ = cot ²θ₁ + cot ²θ₂.

If θ 1 and θ 2 be the apparent angles of dip observed in two vertical… | NEET 2017

Q11NEET 2017Medium

Earth's Magnetism

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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Answer: (d)

In perpendicular planes: $\tan\theta_1 = \frac{V}{H\cos\alpha}$ , $\tan\theta_2 = \frac{V}{H\sin\alpha}$ .
$\cot^2\theta_1 + \cot^2\theta_2 = \frac{H^2(\cos^2\alpha + \sin^2\alpha)}{V^2} = \frac{H^2}{V^2} = \cot^2\theta$
So the answer is (d) $\cot^2\theta = \cot^2\theta_1 + \cot^2\theta_2$ .

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