--- title: "Resection point P from L, M, N" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:resection-lmn-p-2079j" source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-lmn-p-2079j site: BCT Engineering --- # Resection point P from L, M, N Surveying II · Chapter 5: Orientation - Asked in: 2079 Jestha, 6 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-lmn-p-2079j - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question Station L, M and N have the following respective coordinates (2880.24 mE, 8760.12 mN), (3820.60 mE, 8000.25 mN) and (3010.40 mE, 7588.80 mN) respectively. Station P is resection point and following observations were recorded. Determine the coordinate of P. | Stn | Sighted to | HCR | |---|---|---| | P | L | 90°00'00" | | | M | 230°58'51" | | | N | 313°17'05" | | | L | 90°00'30" | ## Short answer Using Tienstra's method with observed angles at P adjusted to ∠LPM=140°58'41", ∠MPN=82°18'04", ∠NPL=136°43'15", and triangle angles computed from coordinates, the resection point coordinates are P (3170.170 mE, 8252.881 mN). ## Answer The three-point problem is solved here by Tienstra's method. #### Observed angles at P Readings: L = 90°00'00", M = 230°58'51", N = 313°17'05", L (closing) = 90°00'30". - $\angle LPM = 230^\circ58'51'' - 90^\circ00'00'' = 140^\circ58'51''$ - $\angle MPN = 313^\circ17'05'' - 230^\circ58'51'' = 82^\circ18'14''$ - $\angle NPL = 90^\circ00'30'' + 360^\circ - 313^\circ17'05'' = 136^\circ43'25''$ The sum is 360°00'30", so the closing error of 30" is shared equally (−10" to each angle): $\angle LPM = 140^\circ58'41''$, $\angle MPN = 82^\circ18'04''$, $\angle NPL = 136^\circ43'15''$. **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: $\alpha = \angle S_2PS_3$, $\beta = \angle S_3PS_1$, $\gamma = \angle S_1PS_2$ ($\alpha+\beta+\gamma = 360^\circ$). 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) | |---|---|---| | L | 2880.240 | 8760.120 | | M | 3820.600 | 8000.250 | | N | 3010.400 | 7588.800 | Angles at P: $\angle LPM = \gamma = 140°58'41"$, $\angle MPN = \alpha = 82°18'04"$, $\angle NPL = \beta = 136°43'15"$ (sum = 360°00'00"). #### Step 1: Angles of the triangle LMN from the coordinates | Side | Bearing | Length (m) | |---|---|---| | LM | 128°56'25" | 1208.999 | | MN | 243°04'37" | 908.689 | | NL | 353°39'33" | 1178.530 | Interior angles: $A$ (at L) $= 44°43'08"$, $B$ (at M) $= 65°51'49"$, $C$ (at N) $= 69°25'04"$ (sum 180°00'00"). #### Step 2: Tienstra constants $$ \begin{aligned} K_1 &= \frac{1}{\cot 44°43' - \cot 82°18'04"} = 1.14328 \\ K_2 &= \frac{1}{\cot 65°51' - \cot 136°43'15"} = 0.66224 \\ K_3 &= \frac{1}{\cot 69°25' - \cot 140°58'41"} = 0.62133 \end{aligned} $$ $\Sigma K = 2.42684$ #### Step 3: Coordinates of P $$ E_P = \frac{(1.14328)(2880.240) + (0.66224)(3820.600) + (0.62133)(3010.400)}{2.42684} = 3170.170\ \text{m} $$ $$ N_P = \frac{(1.14328)(8760.120) + (0.66224)(8000.250) + (0.62133)(7588.800)}{2.42684} = 8252.881\ \text{m} $$ **Check:** the angles computed back from these coordinates are 140°58'41", 82°18'04" and 136°43'15", equal to the observed angles. **Answer: coordinates of P = (3170.170 E, 8252.881 N).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is 360° added when finding the angle NPL from the readings 313°17'05" (N) and 90°00'30" (closing L)? A. Because the closing reading is smaller than the previous reading, so the angle is measured clockwise through zero B. Because the closing error must be removed C. Because the face-left and face-right readings are combined D. Because the angle must always exceed 180° Answer: A. The closing reading on L (90°00'30") is less than the reading on N, so the telescope has passed through 360°. Adding 360° gives the correct clockwise angle of 136°43'25". (Key point: Angle from circle readings) 2. The three angles at P sum to 360°00'30". What is done before applying Tienstra's formula? A. The 30" is added to the largest angle only B. The 30" is ignored since it is small C. The 30" is shared equally, subtracting 10" from each angle D. The readings are repeated with a different formula Answer: C. The angles at P must total exactly 360°, so the 30" excess is divided equally, -10" per angle. (Key point: Distributing the angular misclosure) 3. In Tienstra's formula the constant for station L uses interior angle A (at L) together with which observed angle at P? A. The angle LPM B. The angle MPN, subtended by the side opposite L C. The angle NPL D. The sum of all three angles at P Answer: B. Each K pairs the triangle angle at a station with the angle at P subtended by the opposite side. For L that side is MN, so alpha = angle MPN. (Key point: Pairing K with angles) 4. How are the coordinates of P obtained once K1, K2 and K3 are known? A. As the plain average of the three stations' coordinates B. As the weighted mean E_P = (K1E1 + K2E2 + K3E3)/(K1+K2+K3), and likewise for N C. By adding the K values to the station coordinates D. By taking the coordinates of the station with the largest K Answer: B. The K values act as weights, so P is the weighted mean of the three known stations, with the sum of K in the denominator. 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