Local attraction ABCDEA by included angles

Because the FB and BB of every line differ from 180° by 1°30′ to 3°00′, all the stations (A, B, C, D, E) are affected by local attraction (no line is free of it). The included angle method is therefore used, since an angle at a station is not affected by local attraction at that station.

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.

• A: FB of AB − BB of EA = 191°30′ − 53°00′ = 138°30′
• B: FB of BC − BB of AB = 69°30′ − 13°00′ = 56°30′
• C: 360° − (BB of BC − FB of CD) = 360° − (246°30′ − 32°15′) = 145°45′
• D: FB of DE − BB of CD = 262°45′ − 210°30′ = 52°15′
• E: FB of EA − BB of DE = 230°15′ − 80°45′ = 149°30′

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

Sum observed = 542°30′; error = +2°30′; correction per angle = −0°30′ (distributed equally, since each angle is measured with the same care).

No line has a difference of exactly 180°. Line AB has the smallest discrepancy (1°30′), so it is taken as the most reliable; its adopted bearing is the mean of the FB and (BB + 180°): (191°30′ + 193°00′)/2 = 192°15′.

Step 3: Corrected bearings from the corrected angles

Rule used (traverse run anticlockwise): FB of next line = BB of previous line + interior angle (subtract 360° if the result exceeds 360°).

• BC: 12°15′ + 56°00′ = 68°15′
• CD: 248°15′ + 145°15′ − 360° = 33°30′
• DE: 213°30′ + 51°45′ = 265°15′
• EA: 85°15′ + 149°00′ = 234°15′
• Check, AB: 54°15′ + 138°00′ = 192°15′ (agrees with the adopted value)

Step 4: Corrected bearings

Check: corrected FB and corrected BB differ by exactly 180° for every line.

Answer: all stations A to E are affected. Corrected included angles: A = 138°00′, B = 56°00′, C = 145°15′, D = 51°45′, E = 149°00′ (sum 540°). The corrected bearings are given in the last table (FB: AB 192°15′, BC 68°15′, CD 33°30′, DE 265°15′, EA 234°15′).

Assumption: no line in the traverse has FB and BB differing by exactly 180°, so the line with the smallest discrepancy (AB) is taken as the reference and its mean bearing is adopted.