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Corrected bearings and interior angles, PQRST

Short answer

Corrected fore bearings: PQ = S 37°30′ E, QR = S 43°33′ W, RS = N 73°24′ W, ST = N 11°54′ E, TP = N 58°57′ E. Interior angles: P = 96°27′, Q = 98°57′, R = 116°57′, S = 94°42′, T = 132°57′.

The following fore and back bearings were observed while traversing in an area with compass.

LineF.B.B.B.
PQS 37°30′ EN 37°30′ W
QRS 43°15′ WN 44°15′ E
RSN 73°00′ WS 72°15′ E
STN 12°45′ ES 13°15′ W
TPN 60°00′ ES 59°15′ W

Find the corrected bearings of the lines and also the interior angles of the traverse.

The bearings are converted to whole circle bearings (PQ: FB = S37°30′E = 142°30′, BB = N37°30′W = 322°30′; QR: 223°15′ and 44°15′; RS: 287°00′ and 107°45′; ST: 12°45′ and 193°15′; TP: 60°00′ and 239°15′). Line PQ has FB − BB = 180°00′, so P and Q are free. The other lines differ by −1°00′ (QR), −0°45′ (RS), −0°30′ (ST) and +0°45′ (TP), so stations R, S and T are suspected. Station corrections worked from the free line do not close (the correction found for P comes out as −1°30′ although P 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, FB−BB=180∘\text{FB} - \text{BB} = 180^\circ exactly.

Step 1: Check each line

LineFBBBFB − BBRemark
PQS 37°30′ EN 37°30′ W180°00′Free, difference = 180°
QRS 43°15′ WN 44°15′ E179°00′Differs by −1°00′
RSN 73°00′ WS 72°15′ E179°15′Differs by −0°45′
STN 12°45′ ES 13°15′ W179°30′Differs by −0°30′
TPN 60°00′ ES 59°15′ W180°45′Differs by +0°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
PPT and PQ96°45′96°27′
QQP and QR99°15′98°57′
RRQ and RS117°15′116°57′
SSR and ST95°00′94°42′
TTS and TP133°15′132°57′

Sum observed = 541°30′; error = +1°30′; correction per angle = −0°18′ (equal distribution, since each angle is measured with the same care).

Line PQ 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 (in WCB): FB of next line = BB of previous line − interior angle (traverse run clockwise). Starting from PQ = 142°30′ (BB 322°30′): QR = 322°30′ − 98°57′ = 223°33′ = S 43°33′ W, and so on round the traverse.

LineObserved FBCorrectionCorrected FBObserved BBCorrected BB
PQS 37°30′ E0S 37°30′ EN 37°30′ WN 37°30′ W
QRS 43°15′ W+0°18′S 43°33′ WN 44°15′ EN 43°33′ E
RSN 73°00′ W−0°24′N 73°24′ WS 72°15′ ES 73°24′ E
STN 12°45′ E−0°51′N 11°54′ ES 13°15′ WS 11°54′ W
TPN 60°00′ E−1°03′N 58°57′ ES 59°15′ WS 58°57′ W

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

Corrections are in the whole-circle sense (+ is clockwise).

Interior angles from the corrected bearings

StationInterior angle
P96°27′
Q98°57′
R116°57′
S94°42′
T132°57′

Sum = 540°00′ (as required).

Answer: Corrected bearings (FB): PQ = S 37°30′ E, QR = S 43°33′ W, RS = N 73°24′ W, ST = N 11°54′ E, TP = N 58°57′ E. Interior angles: P = 96°27′, Q = 98°57′, R = 116°57′, S = 94°42′, T = 132°57′.

✓ Answer checked by Claude Opus 5.5 (high effort)

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