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If the reduced bearing of a line AB is N60°W and length is 100 m, then the latitude and departure respectively of the line AB will be
  • a)
    + 50 m, + 86.6 m
  • b)
    + 86.6 m , - 5 0 m
  • c)
    + 50 m, - 86 . 6 m 
  • d)
    + 70.7 m, - 50 m
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If the reduced bearing of a line AB is N60°W and length is 100 m, ...
E, the whole circle bearing of AB would be S30W.

To determine the whole circle bearing, you need to add 180 degrees to the reduced bearing if it falls in the first or second quadrant, or subtract 180 degrees if it falls in the third or fourth quadrant.

In this case, N60E falls in the northeast quadrant, so you add 180 degrees to get the whole circle bearing:

N60E + 180 = S120E

However, the convention is to express bearings as angles measured clockwise from north, so you need to subtract the whole circle bearing from 360 degrees:

360 - 120 = 240

Therefore, the whole circle bearing of line AB is S30W or 240 degrees.
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Community Answer
If the reduced bearing of a line AB is N60°W and length is 100 m, ...
Lcos∆=100cos60=50
dsin∆=100sin60= -86.6

N60W.... LATITUDE IS POSITIVE
DEPARTURE IS NEGATIVE
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If the reduced bearing of a line AB is N60°W and length is 100 m, then the latitude and departure respectively of the line AB will bea)+ 50 m, + 86.6 mb)+ 86.6 m , - 5 0 mc)+ 50 m, - 86 . 6 md)+ 70.7 m, - 50 mCorrect answer is option 'C'. Can you explain this answer?
Question Description
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