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Example:
The following fore bearings were observed respectively for lines, AB, BC, CD, DE, EF and FG. Find out their back bearings:
(i) 148°
(ii) 65°
(iii) 285°
(iv) 215°
(v) N 36° W
(vi) S 40° E
Solution: The difference between fore bearing and the back bearing of a line ought to be 180°. Noting that in WCB angle is from 0° to 360°, we find back bearing = fore bearing ± 180° + 180° is used if θ is less than 180° and - 180° is used when θ is more than 180°. So
(i) BB of AB = 145° + 180° = 325°
(ii) BB of BC = 65° + 180° = 245°
(iii) BB of CD = 285° - 180° = 105°
(iv) BB of DE = 215° - 180° = 35°
In particular case of RB, back bearing of a line may be obtained by interchanging N and S at the identical time E and W. so
(v) BB of EF = S 36° E (vi) BB of FG = N 40° W.
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