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Question
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 , `δ_1 = 45^circ`
Dip shown by the needle of the dip circle on rotating the circle, `δ_2 = 53^circ`
true dip (δ) is given by
`cot^2 δ = cot^2 δ_1 + cot^2 δ_2`
⇒`cot^2 δ = cot^2 45^circ + cot^2 53^circ`
⇒`cot^2 δ = 1.56`
⇒`cot^2 δ = 1.56`
⇒`δ = 38.6^circ ≈ 39^circ`
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