A magnetic needle is free to rotate in a vertical plane, which makes an angle of $$600$$ with the magnetic meridian. If the needle stays in a direction making an angle of $$tan$$$$-123$$ with the horizontal , what would be the dip at that place?

Question

A magnetic needle is free to rotate in a vertical plane, which makes an angle of $$600$$ with the magnetic meridian. If the needle stays in a direction making an angle of $$tan$$$$-123$$ with the horizontal , what would be the dip at that place?

Infinity Learn · Accepted Answer

We know the relationship between true dip and apparent dip        $$tan$$δ$$1=$$$$tan$$δ$$cos$$θGiven : Apparent dip δ$$1=$$$$tan$$$$-123$$ ⇒$$tan$$δ$$1$$ $$=$$ $$23$$   True dip  δ$$=$$?                         ALso , θ$$=60$$°∴$$23=$$$$tan$$δ$$cos60$$°⇒$$tan$$δ$$=13$$⇒$$tan$$δ$$=tan30$$°⇒δ$$=30$$°

A magnetic needle is free to rotate in a vertical plane, which makes an angle of $60^{0}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{- 1} (\frac{2}{\sqrt{3}})$ with the horizontal , what would be the dip at that place?

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A magnetic needle is free to rotate in a vertical plane, which makes an angle of 600 with the magnetic meridian. If the needle stays in a direction making an angle of tan-123 with the horizontal , what would be the dip at that place?

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A magnetic needle is free to rotate in a vertical plane, which makes an angle of $60^{0}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{- 1} (\frac{2}{\sqrt{3}})$ with the horizontal , what would be the dip at that place?