The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.

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Solution

Given:
Apparent dip shown by the needle of the dip circle, δ₁ = 45°
Dip shown by the needle of the dip circle on rotating the circle, δ₂= 53°
True dip $(δ)$ is given by
Cot² δ = Cot² δ₁ + Cot² δ₂
$\Rightarrow$ Cot²δ = Cot² 45° + Cot² 53°
$\Rightarrow$ Cot²δ = 1.56
$\Rightarrow$ Cot²δ = 1.56
$\Rightarrow$ δ = 38.6° ≈ 39°

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