Interior angles and bearings, closed traverse ABCD

Line AB has FB − BB = 180°00′, so stations A and B are free; stations C and D are corrected starting from it.

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

Step 2: Correct the stations in turn

• Line BC: FB at B = 48°00′ is correct. The correct BB is 228°00′, observed 226°00′, so the correction at C is +2°00′.
• Line CD: FB at C = 290°30′ with correction +2°00′ gives 292°30′. The correct BB is 112°30′, observed 115°15′, so the correction at D is −2°45′.
• Line DA (check): FB at D = 180°15′ with correction −2°45′ gives 177°30′. The correct BB is 357°30′, which equals the observed value, so A needs no correction, as expected.

Stations affected by local attraction: C, D.

Step 3: Corrected bearings

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

Interior angles

The included angle at a station is found from the observed bearings taken at that station (FB of the leaving line minus BB of the arriving line), so local attraction does not affect it:

Sum = 360°00′ = $(2\times4-4)\times 90^\circ$, so the angles need no correction. They agree with the angles obtained from the corrected bearings.

Answer: Interior angles A = 107°00′, B = 123°30′, C = 64°30′, D = 65°00′ (no correction needed). Corrected bearings (FB): AB = 104°30′, BC = 48°00′, CD = 292°30′, DA = 177°30′.