Local attraction, bearings AB 305°30′

Line AB has FB − BB = 180°00′, so stations A and B are free. The differences for the other lines are BC +1°00′, CD −2°00′, DE +1°30′ and EA +1°00′, so C, D and E are suspected (A and B are not). Correcting the stations one by one starting from AB does not close: going round the traverse, the correction found for A comes out as +1°30′ although A is free. 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 = 538°30′; error = −1°30′; correction per angle = +0°18′ (equal distribution, since each angle is measured with the same care).

Line AB 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 clockwise). Starting from AB = 305°30′ (BB 125°30′): BC = 125°30′ − 50°18′ = 75°12′, 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 = 305°30′, BC = 75°12′, CD = 115°54′, DE = 164°36′, EA = 224°18′.