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If the magnetic bearing of the Sun at a place at noon is S 2° E. the magnetic declination (in degrees) at that place is
  • a)
    2° W
  • b)
    4° E
  • c)
    4° W
  • d)
    2° E
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
If the magnetic bearing of the Sun at a place at noon is S 2° E. t...
8° W, it means that the angle between the direction of true north and the direction towards which the north-seeking end of a magnetic needle points is 28° towards the west. This is known as the magnetic declination or variation at that place.

To find the true bearing of the Sun at that place, we need to add or subtract the magnetic declination from the magnetic bearing, depending on whether the declination is east or west.

If the declination is west, as in this case, we need to subtract it from the magnetic bearing to get the true bearing. Therefore, the true bearing of the Sun at that place at noon would be:

True bearing = Magnetic bearing - Magnetic declination
True bearing = S 28° W - 28° W
True bearing = S 0° W

This means that the direction towards which the Sun appears to be at noon is towards true south, i.e., directly overhead at the equator.
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Community Answer
If the magnetic bearing of the Sun at a place at noon is S 2° E. t...
Magnetic bearing (MB) of Sun in WCB = 178°.
True bearing (TB) of Sun in WCB at noon must be 180°.
MB + δ = TB
178° + δ = 180°
δ = 2°
∵ Declination (δ) is found to be positive so it is East declination of 2° i.e. 2°E.
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If the magnetic bearing of the Sun at a place at noon is S 2° E. the magnetic declination (in degrees) at that place isa)2° Wb)4° Ec)4° Wd)2° ECorrect answer is option 'D'. Can you explain this answer?
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