Local attraction, S 40°30′ W traverse

The bearings are converted to whole circle bearings: AB = S40°30′W = 220°30′ (BB 41°15′), BC = 260°45′ (BB 79°30′), CD = 19°30′ (BB 200°00′), DA = S80°00′E = 100°00′ (BB 280°00′). Line DA has FB − BB = 180°00′, so D and A are free; B and C are suspected.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Line DA has a difference of exactly 180°, so stations D and A are free from local attraction. The other stations are corrected one after another, starting from A.

Step 2: Correct the stations in turn

• Line AB: FB at A = S 40°30′ W is correct. The correct BB is N 40°30′ E, observed N 41°15′ E, so the correction at B is −0°45′.
• Line BC: FB at B = S 80°45′ W with correction −0°45′ gives S 80°00′ W. The correct BB is N 80°00′ E, observed N 79°30′ E, so the correction at C is +0°30′.
• Line CD (check): FB at C = N 19°30′ E with correction +0°30′ gives N 20°00′ E. The correct BB is S 20°00′ W, which equals the observed value, so D needs no correction, as expected.

Stations affected by local attraction: B, C.

Step 3: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Corrections are in the whole-circle sense (+ is clockwise).

Interior angles from the corrected bearings

Sum = 360°00′ (as required).

Answer: Local attraction is suspected at B and C. Corrected bearings (FB): AB = S 40°30′ W, BC = S 80°00′ W, CD = N 20°00′ E, DA = S 80°00′ E.

The interior angles computed from the observed bearings are A = 59°30′, B = 140°30′, C = 60°00′, D = 100°00′ (sum 360°00′), the same as those from the corrected bearings.