--- title: "Coordinates and bearings of P from A, B, C" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:resection-apb-bpc-lengths-bearings" source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-apb-bpc-lengths-bearings site: BCT Engineering --- # Coordinates and bearings of P from A, B, C Surveying II · Chapter 5: Orientation - Asked in: 2072 Asoj, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-apb-bpc-lengths-bearings - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question The following are the co-ordinates of three known station points whose directions are observed from the unknown instrument station P. | | A | B | C | |---|---|---|---| | Easting (m) | 5,000 | 10,000 | 15,000 | | Northing (m) | 10,000 | 15,000 | 10,000 | If observed horizontal angle APB = 45° and BPC = 52° Determine (i) Length and Bearings of AP, BP and CP (ii) Co-ordinates of P. ## Short answer With corrected Northing of A = 1000 m, Tienstra's method gives P = (18900.471 mE, 15046.510 mN). Lengths and bearings: AP = 19761.769 m (44°42'02"), BP = 8900.592 m (89°42'02"), CP = 6378.161 m (37°42'02"). ## Answer **Note on the data.** With A = (5000 E, 10000 N), B = (10000, 15000) and C = (15000, 10000), triangle ABC is right-angled at B and isosceles ($\angle A = \angle C = 45^\circ$). The points that see AB at 45° on C's side of AB lie on the circle through A, B and C, and from every such point BC is seen at 45° or 135°, never 52°. Checking all other positions (including the far side of AB) also gives no point where $\angle APB = 45^\circ$ and $\angle BPC = 52^\circ$ at the same time. So the data as printed has no solution. The northing of A is therefore taken as 1000.000 m (as in the companion problem), which gives a proper resection. P then lies outside the triangle. **Angles at P:** $\angle APB = 45^\circ$, $\angle BPC = 52^\circ$; the rays are in the order C, A, B, so $\angle CPA = 52^\circ - 45^\circ = 7^\circ$. #### Triangle ABC from coordinates (E, N) | Side | Bearing | Length (m) | |---|---|---| | AB | 19°39'14" | 14866.069 | | BC | 135°00'00" | 7071.068 | | CA | 228°00'46" | 13453.624 | Interior angles: $A = 28°21'32"$, $B = 64°39'14"$, $C = 86°59'14"$. #### Tienstra's method (directed angles: $\gamma = \angle APB = +45^\circ$, $\alpha = \angle BPC = -52^\circ$, $\beta = \angle CPA = +7^\circ$) $$ K_A = \frac{1}{\cot A - \cot\alpha} = 0.37966,\quad K_B = \frac{1}{\cot B - \cot\beta} = -0.13037,\quad K_C = \frac{1}{\cot C - \cot\gamma} = -1.05556 $$ $$ E_P = \frac{\sum K E}{\sum K} = 18900.471\ \text{m},\qquad N_P = \frac{\sum K N}{\sum K} = 15046.510\ \text{m} $$ #### (i) Lengths and bearings $$ \text{Length} = \sqrt{\Delta E^2 + \Delta N^2}, \qquad \text{bearing} = \tan^{-1}\frac{\Delta E}{\Delta N}\ (\text{quadrant corrected}) $$ | Line (from P) | $\Delta E$ (m) | $\Delta N$ (m) | Length (m) | Bearing from P | Bearing towards P | |---|---|---|---|---|---| | PA | -13900.471 | -14046.510 | 19761.769 | 224°42'02" | AP = 44°42'02" | | PB | -8900.471 | -46.510 | 8900.592 | 269°42'02" | BP = 89°42'02" | | PC | -3900.471 | -5046.510 | 6378.161 | 217°42'02" | CP = 37°42'02" | Check: bearing PB − bearing PA $= 45°00'00"$ and bearing PB − bearing PC $= 52°00'00"$, equal to the observed angles. The bearings of AP, BP and CP (from the stations to P) are the bearings from P $\pm 180^\circ$. #### (ii) Coordinates of P **Answer (with $N_A$ = 1000 m): P = (18900.471 E, 15046.510 N). AP = 19761.769 m (bearing 44°42'02"), BP = 8900.592 m (bearing 89°42'02"), CP = 6378.161 m (bearing 37°42'02").** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the answer change the northing of A to 1000.000 m? A. The original data would make P lie on the danger circle through A, B, C, so no true solution exists B. A is too far from B C. The angles are too small D. To simplify arithmetic Answer: A. With A at 10000 N, angle ACB = 45° equals the observed APB, so P would be on the danger circle and BPC could not be 52°. (Key point: Data inconsistency) 2. The rays from P are in order C, A, B with APB = 45° and BPC = 52°. What is angle CPA? A. 97° B. 7° C. 263° D. 45° Answer: B. P lies outside, so angle BPC includes angle CPA and APB: CPA = 52° - 45° = 7°. (Key point: Third angle at P) 3. Why are directed angles gamma = +45°, alpha = -52°, beta = +7° used in Tienstra's formula here? A. Because P is outside the triangle and the angles are not all in the same turning sense B. Because angles are measured anticlockwise C. To make the sum 180° D. The signs have no meaning Answer: A. For an outside point the angles do not add to 360° in the same sense, so signs are assigned, and these still sum to 0 for directed angles. (Key point: Signed (directed) angles) 4. How is the result for the lengths and bearings verified? A. Bearing PB - bearing PA = 45° and bearing PB - bearing PC = 52° B. All three lengths are equal C. Bearings sum to 360° D. PA + PB = PC Answer: A. The angle between two rays equals the difference of their bearings: 269°42'02" - 224°42'02" = 45° and 269°42'02" - 217°42'02" = 52°. 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