Corrected bearings and included angles (65)

Line DE has FB − BB = 180°00′, so stations D and E are free. The other lines show errors (AB −1°30′, BC +3°00′, CD +1°45′, EA −2°45′), so A, B and C are suspected. Line EA differs by −2°45′ although E is free (by DE), which shows that the bearings also contain observational errors and not only local attraction. The station corrections do not agree when worked in the two directions round the traverse, so the included-angle method is used.

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

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

The sum of the interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 540°30′; error = +0°30′; correction per angle = −0°06′ (equal distribution, since each angle is measured with the same care).

Line DE has FB − BB = 180°, so its bearing is free from local attraction and is taken as the starting line.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise). Starting from DE = 262°45′ (BB 82°45′): EA = 82°45′ + 147°24′ = 230°09′, and so on round the traverse.

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

Interior angles from the corrected bearings

Sum = 540°00′ (as required).

Answer: Corrected bearings (FB): AB = 188°33′, BC = 64°57′, CD = 30°36′, DE = 262°45′, EA = 230°09′. Included angles: A = 138°24′, B = 56°24′, C = 145°39′, D = 52°09′, E = 147°24′.