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Local attraction, BC 79°30′ and DE 82°15′

Short answer

No line is free; all stations suspected. Included-angle method: interior angles sum 541°00′, error +1°00′, correction −0°12′ per angle. The most reliable line DE (smallest diff) is adopted as 262°30′. Corrected fore bearings: AB 188°36′, BC 74°54′, CD 30°27′, DE 262°30′, EA 230°18′.

Following are the bearings observed in a compass traverse survey. At what stations do you suspect local attraction? Correct them by applying a suitable correction method.

LineFBBB
AB191°30′13°00′
BC79°30′256°30′
CD32°15′210°30′
DE262°45′82°15′
EA230°15′53°00′

No line is free (the differences are 178°30′, 183°00′, 181°45′, 180°30′ and 177°15′), so all stations are suspected of local attraction. Line DE has the smallest difference (0°30′), so D and E are the least affected; 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, FB−BB=180∘\text{FB} - \text{BB} = 180^\circ exactly.

Step 1: Check each line

LineFBBBFB − BBRemark
AB191°30′13°00′178°30′Differs by −1°30′
BC79°30′256°30′183°00′Differs by +3°00′
CD32°15′210°30′181°45′Differs by +1°45′
DE262°45′82°15′180°30′Differs by +0°30′
EA230°15′53°00′177°15′Differs by −2°45′

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)×90∘=540∘(2n-4)\times 90^\circ = 540^\circ.

StationAngle betweenObservedCorrected
AAE and AB138°30′138°18′
BBA and BC66°30′66°18′
CCB and CD135°45′135°33′
DDC and DE52°15′52°03′
EED and EA148°00′147°48′

Sum observed = 541°00′; error = +1°00′; correction per angle = −0°12′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line DE has the smallest difference (+0°30′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 262°30′.

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°30′ (BB 82°30′): EA = 82°30′ + 147°48′ = 230°18′, and so on round the traverse.

LineObserved FBCorrectionCorrected FBObserved BBCorrected BB
AB191°30′−2°54′188°36′13°00′8°36′
BC79°30′−4°36′74°54′256°30′254°54′
CD32°15′−1°48′30°27′210°30′210°27′
DE262°45′−0°15′262°30′82°15′82°30′
EA230°15′+0°03′230°18′53°00′50°18′

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

Interior angles from the corrected bearings

StationInterior angle
A138°18′
B66°18′
C135°33′
D52°03′
E147°48′

Sum = 540°00′ (as required).

Answer: Corrected bearings (FB): AB = 188°36′, BC = 74°54′, CD = 30°27′, DE = 262°30′, EA = 230°18′.

✓ Answer checked by Claude Opus 5.5 (high effort)

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