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A magnet suspended at `30^@` with magnetic meridian makes an angle of `45^@` with the horizontal. What shall be the actual value of the angle of dip? |
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Answer» `tan 45^(@) = (tan delta)/(cos 30^(@))` `1 = (2 tan delta)/(sqrt3)` `delta = tan^(-1) ((sqrt3)/(2))` |
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