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.