- 2066 Bhadra · 9 marks
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.
| Line | F.B. | B.B. |
|---|---|---|
| PQ | S 37°30′ E | N 37°30′ W |
| QR | S 43°15′ W | N 44°15′ E |
| RS | N 73°00′ W | S 72°15′ E |
| ST | N 12°45′ E | S 13°15′ W |
| TP | N 60°00′ E | S 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, exactly.
Step 1: Check each line
| Line | FB | BB | FB − BB | Remark |
|---|---|---|---|---|
| PQ | S 37°30′ E | N 37°30′ W | 180°00′ | Free, difference = 180° |
| QR | S 43°15′ W | N 44°15′ E | 179°00′ | Differs by −1°00′ |
| RS | N 73°00′ W | S 72°15′ E | 179°15′ | Differs by −0°45′ |
| ST | N 12°45′ E | S 13°15′ W | 179°30′ | Differs by −0°30′ |
| TP | N 60°00′ E | S 59°15′ W | 180°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 .
| Station | Angle between | Observed | Corrected |
|---|---|---|---|
| P | PT and PQ | 96°45′ | 96°27′ |
| Q | QP and QR | 99°15′ | 98°57′ |
| R | RQ and RS | 117°15′ | 116°57′ |
| S | SR and ST | 95°00′ | 94°42′ |
| T | TS and TP | 133°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.
| Line | Observed FB | Correction | Corrected FB | Observed BB | Corrected BB |
|---|---|---|---|---|---|
| PQ | S 37°30′ E | 0 | S 37°30′ E | N 37°30′ W | N 37°30′ W |
| QR | S 43°15′ W | +0°18′ | S 43°33′ W | N 44°15′ E | N 43°33′ E |
| RS | N 73°00′ W | −0°24′ | N 73°24′ W | S 72°15′ E | S 73°24′ E |
| ST | N 12°45′ E | −0°51′ | N 11°54′ E | S 13°15′ W | S 11°54′ W |
| TP | N 60°00′ E | −1°03′ | N 58°57′ E | S 59°15′ W | S 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
| Station | Interior angle |
|---|---|
| P | 96°27′ |
| Q | 98°57′ |
| R | 116°57′ |
| S | 94°42′ |
| T | 132°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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