--- title: "Tienstra's method, P from A, B, C (APB=45°)" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:tienstra-resection-p-abc-45-52" source: https://www.bctengineer.com/q/semester_4/surveying_ii/tienstra-resection-p-abc-45-52 site: BCT Engineering --- # Tienstra's method, P from A, B, C (APB=45°) Surveying II · Chapter 5: Orientation - Asked in: 2080 Chaitra, 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/tienstra-resection-p-abc-45-52 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question The followings are the coordinates of their known stations whose direction are observed from unfixed instrument station P. | Stations | A | B | C | |---|---|---|---| | Easting (m) | 5000.000 | 10000.000 | 15000.000 | | Northing (m) | 1000.000 | 15000.000 | 10000.000 | If observed horizontal angle APB = 45°00'00" and angle BPC = 52°00'00", determine the coordinates of resection point P by Tienstra method. ## Short answer Using Tienstra's method with signed observed angles (+45°, -52°, +7°) and interior angles of triangle ABC from coordinates, the computed constants yield P = (18900.471 m E, 15046.510 m N). ## Answer The northing of A is taken as printed, 1000.000 m. P lies outside the triangle ABC. The rays from P are in the order C, A, B: $\angle APB = 45^\circ$, $\angle BPC = 52^\circ$, and therefore $\angle CPA = 52^\circ - 45^\circ = 7^\circ$ (ray PA lies between PC and PB). Taken as directed angles (clockwise positive), they are $+45^\circ$, $-52^\circ$ and $+7^\circ$ (sum 0). This is the only position of P that fits the observations. **Tienstra's formula.** Let the known stations be $S_1, S_2, S_3$ with interior angles $A, B, C$ of the triangle $S_1S_2S_3$ at them. Let $\alpha, \beta, \gamma$ be the observed angles at P subtended by the opposite sides: $\alpha = \angle S_2PS_3$, $\beta = \angle S_3PS_1$, $\gamma = \angle S_1PS_2$ ($\alpha+\beta+\gamma = 360^\circ$, or $0^\circ$ when directed angles of mixed sign are used, as here). Then $$ K_1 = \frac{1}{\cot A - \cot\alpha},\quad K_2 = \frac{1}{\cot B - \cot\beta},\quad K_3 = \frac{1}{\cot C - \cot\gamma} $$ $$ E_P = \frac{K_1E_1 + K_2E_2 + K_3E_3}{K_1+K_2+K_3}, \qquad N_P = \frac{K_1N_1 + K_2N_2 + K_3N_3}{K_1+K_2+K_3} $$ #### Data | Station | Easting (m) | Northing (m) | |---|---|---| | A | 5000.000 | 1000.000 | | B | 10000.000 | 15000.000 | | C | 15000.000 | 10000.000 | Angles at P: $\angle APB = \gamma = +45°00'00"$, $\angle BPC = \alpha = -52°00'00"$, $\angle CPA = \beta = +7°00'00"$ (sum = 0°00'00"). #### Step 1: Angles of the triangle ABC from the coordinates | Side | Bearing | Length (m) | |---|---|---| | AB | 19°39'14" | 14866.069 | | BC | 135°00'00" | 7071.068 | | CA | 228°00'46" | 13453.624 | Interior angles: $A$ (at A) $= 28°21'32"$, $B$ (at B) $= 64°39'14"$, $C$ (at C) $= 86°59'14"$ (sum 180°00'00"). #### Step 2: Tienstra constants $$ \begin{aligned} K_1 &= \frac{1}{\cot 28°21'32" - \cot(-52^\circ)} = 0.37966 \\ K_2 &= \frac{1}{\cot 64°39'14" - \cot 7^\circ} = -0.13037 \\ K_3 &= \frac{1}{\cot 86°59'14" - \cot 45^\circ} = -1.05556 \end{aligned} $$ $\Sigma K = -0.80626$ #### Step 3: Coordinates of P $$ E_P = \frac{(0.37966)(5000.000) + (-0.13037)(10000.000) + (-1.05556)(15000.000)}{-0.80626} = 18900.471\ \text{m} $$ $$ N_P = \frac{(0.37966)(1000.000) + (-0.13037)(15000.000) + (-1.05556)(10000.000)}{-0.80626} = 15046.510\ \text{m} $$ **Check:** the clockwise angles computed back from these coordinates are A to B = 45°00'00", B to C = 308°00'00" (that is, 52° anticlockwise) and C to A = 7°00'00", agreeing with the observations. **Answer: coordinates of P = (18900.471 E, 15046.510 N).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With P outside triangle ABC, ∠APB = 45° and ∠BPC = 52°, what is ∠CPA and why? A. 83°, as 360° - 45° - 52° - 180° B. 97°, as 45° + 52° C. 7°, as ray PA lies between PC and PB so 52° - 45° D. 263°, as 360° - 97° Answer: C. The rays from P are in the order C, A, B, so ∠CPA = 52° - 45° = 7°. (Key point: Finding the third angle) 2. When the angles are taken as directed (clockwise positive), what do the three angles at P come to? A. +45°, -52° and +7°, summing to 0° B. +45°, +52° and +7°, summing to 104° C. +45°, +52° and +263°, summing to 360° D. -45°, -52° and -7°, summing to -104° Answer: A. Since P is outside the triangle the directed angles sum to 0° instead of 360°. (Key point: Signed angles) 3. Here K2 and K3 and ΣK are negative. What does that mean for the computation? A. The method fails and a different method must be used B. The signs are dropped and absolute values are used C. Only the stations with positive K are used D. The same weighted-mean formula is still used; the signs come from P lying outside the triangle Answer: D. The formula E = ΣKE/ΣK still holds; the negative constants simply reflect that P is outside the triangle, giving P = (18900.471 E, 15046.510 N). 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