--- title: "Coordinates of instrument station O and station A" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:resection-pwd-instrument-station-a" source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-pwd-instrument-station-a site: BCT Engineering --- # Coordinates of instrument station O and station A Surveying II · Chapter 5: Orientation - Asked in: 2079 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/resection-pwd-instrument-station-a - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question The following are the three known points. The information regarding the three points and one traverse leg are given below. Compute the coordinates of instrument station and the traverse station A. | Instrument station | Sighted to | CW observed angles | Northing (m) | Easting (m) | Remarks | |---|---|---|---|---|---| | O | P | ∠POW = 140°58'51" | 8760.12 | 2880.24 | | | | W | ∠WOD = 82°18'14" | 8000.25 | 3820.60 | | | | A | ∠WOA = 42°18'14" | ? | ? | Length (OA) = 140.23 m | | | D | ∠DOP = 136°42'55" | 7588.80 | 3010.40 | | | | P | | 8760.12 | 2880.24 | | ## Short answer O is determined by Tienstra's method as (3170.205 m E, 8252.871 m N). From bearing OW and angle WOA, bearing OA is computed; using length OA=140.23 m, A = (3232.708 m E, 8127.341 m N). ## Answer The coordinates are written as (Easting, Northing): P = (2880.24 E, 8760.12 N), W = (3820.60 E, 8000.25 N), D = (3010.40 E, 7588.80 N). Observed clockwise angles at O: $\angle POW = 140^\circ58'51''$, $\angle WOD = 82^\circ18'14''$, $\angle DOP = 136^\circ42'55''$ (sum 360°). #### Part 1: Coordinates of the instrument station O (Tienstra's method) **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) | |---|---|---| | P | 2880.240 | 8760.120 | | W | 3820.600 | 8000.250 | | D | 3010.400 | 7588.800 | Angles at O: $\angle POW = \gamma = 140°58'51"$, $\angle WOD = \alpha = 82°18'14"$, $\angle DOP = \beta = 136°42'55"$ (sum = 360°00'00"). #### Step 1: Angles of the triangle PWD from the coordinates | Side | Bearing | Length (m) | |---|---|---| | PW | 128°56'25" | 1208.999 | | WD | 243°04'37" | 908.689 | | DP | 353°39'33" | 1178.530 | Interior angles: $A$ (at P) $= 44°43'08"$, $B$ (at W) $= 65°51'49"$, $C$ (at D) $= 69°25'04"$ (sum 180°00'00"). #### Step 2: Tienstra constants $$ \begin{aligned} K_1 &= \frac{1}{\cot 44°43' - \cot 82°18'14"} = 1.14321 \\ K_2 &= \frac{1}{\cot 65°51' - \cot 136°42'55"} = 0.66233 \\ K_3 &= \frac{1}{\cot 69°25' - \cot 140°58'51"} = 0.62128 \end{aligned} $$ $\Sigma K = 2.42682$ #### Step 3: Coordinates of O $$ E_O = \frac{(1.14321)(2880.240) + (0.66233)(3820.600) + (0.62128)(3010.400)}{2.42682} = 3170.205\ \text{m} $$ $$ N_O = \frac{(1.14321)(8760.120) + (0.66233)(8000.250) + (0.62128)(7588.800)}{2.42682} = 8252.871\ \text{m} $$ **Check:** the angles computed back from these coordinates are 140°58'51", 82°18'14" and 136°42'55", equal to the observed angles. **Coordinates of O = (3170.205 E, 8252.871 N).** #### Part 2: Coordinates of the traverse station A $\angle WOA = 42^\circ18'14''$ is measured clockwise from OW (A lies between W and D, since $\angle WOD = 82^\circ18'14''$), and $OA = 140.23$ m. Bearing of OW (from O to W) $= 111°13'36"$ Bearing of OA = bearing of OW + 42°18'14" $= 153°31'50"$ $$ E_A = E_O + OA\sin\theta = 3170.205 + 140.23\sin 153°31'50" = 3232.708\ \text{m} $$ $$ N_A = N_O + OA\cos\theta = 8252.871 + 140.23\cos 153°31'50" = 8127.341\ \text{m} $$ **Answer: O = (3170.205 E, 8252.871 N); A = (3232.708 E, 8127.341 N).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the problem split into parts? A. A is found first, then O from A B. O is found from known stations P, W, D by Tienstra, then A is located from O by bearing and distance C. Both are found at once by Tienstra D. O is assumed and A is found by sine rule only Answer: B. Station O is first fixed by three-point resection (Tienstra), then A is set out from O using the bearing of OA and the length OA. (Key point: Step order) 2. Which observed angles at O are used with Tienstra's formula for P, W, D? A. ∠POW, ∠WOD and ∠DOP, which total 360° B. ∠WOA only C. ∠POW and ∠WOA D. ∠WOA and OA length Answer: A. The three angles subtended by the sides of triangle PWD at O (140°58'51", 82°18'14", 136°42'55") sum to 360°; ∠WOA is for locating A. (Key point: Which angles are used for Tienstra) 3. How is the bearing of OA found? A. Bearing of OW minus 42°18'14" B. Bearing of OD plus 42°18'14" C. Bearing of OW plus 42°18'14" (clockwise) = 153°31'50" D. Bearing of OP plus 82°18'14" Answer: C. ∠WOA is measured clockwise from OW, so bearing OA = 111°13'36" + 42°18'14" = 153°31'50". (Key point: Bearing of OA) 4. Which expressions give the coordinates of A? A. E_A = E_O + OA cosθ, N_A = N_O + OA sinθ B. E_A = E_O + OA sinθ, N_A = N_O + OA cosθ C. E_A = E_O - OA sinθ, N_A = N_O - OA cosθ D. E_A = OA sinθ, N_A = OA cosθ Answer: B. With bearing θ measured from north, departure = OA sinθ is added to easting and latitude = OA cosθ is added to northing. 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