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At $45^{o}$ to the magnetic meridian, the apparent dip is $60^{o}$ . The true dip is

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tan-13

tan-113

tan-132

tan-116

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detailed solution

Correct option is C

Apparent dip δ' is given bytan δ'=VH'=VH cos 45o=VH x 1cos 45o=tan δ.2But tan 60o=tan δ.23=tan δ.2tan δ=32 ∴δ=tan-132

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At 45o to the magnetic meridian, the apparent dip is 60o. The true dip is

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At $45^{o}$ to the magnetic meridian, the apparent dip is $60^{o}$ . The true dip is