---
title: "Three point analytical resection, average coordinates of O"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:three-point-resection-p-w-d"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/three-point-resection-p-w-d
site: BCT Engineering
---

# Three point analytical resection, average coordinates of O

Surveying II · Chapter 5: Orientation

- Asked in: 2076 Bhadra, 8 marks
- Video: [Analytical method of Solving three point problem | Collin's method | Surveying | Prashant YT | BE |](https://www.youtube.com/watch?v=i-Af3CXWY4w) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/three-point-resection-p-w-d
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

In a three point analytical resection P, W and D are three known coordinated points and from resection point O, angles observation are taken. Compute the average coordinates of instrument point O from the following data.

| Instrument pointing to | HCR observation | Station | Easting (m) | Northing (m) |
|---|---|---|---|---|
| P | 90°00'00" | P | 2876.240 | 8754.110 |
| W | 230°58'51" | W | 3810.800 | 7997.250 |
| D | 313°17'05" | D | 2959.390 | 7487.090 |
| P | 90°00'30" | | | |

## Short answer

Using Tienstra's formula with observed horizontal angles, three solutions were obtained. Their average gives instrument point O at (3152.804 mE, 8259.788 mN).

## Answer

#### Angles at O

Readings: P = 90°00'00", W = 230°58'51", D = 313°17'05", P (closing) = 90°00'30".

- $\angle POW = 140^\circ58'51''$, $\angle WOD = 82^\circ18'14''$, $\angle DOP$ from the first and the closing reading of P = $136^\circ42'55''$ and $136^\circ43'25''$.

The closing reading differs by 30", so the third angle can be taken in two ways. Each pair of observed angles (the third being 360° minus the sum) gives a solution by Tienstra's formula, and the **average** of the three solutions is taken.

**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

$$
K_1 = \frac{1}{\cot A - \cot\alpha},\quad K_2 = \frac{1}{\cot B - \cot\beta},\quad K_3 = \frac{1}{\cot C - \cot\gamma}
$$

$$
E_P = \frac{K_1E_1 + K_2E_2 + K_3E_3}{K_1+K_2+K_3}, \qquad N_P = \frac{K_1N_1 + K_2N_2 + K_3N_3}{K_1+K_2+K_3}
$$

#### Detailed solution for set 1 ($\angle POW = 140^\circ58'51''$, $\angle WOD = 82^\circ18'14''$, $\angle DOP = 136^\circ42'55''$)


Known stations: P (2876.240 E, 8754.110 N), W (3810.800 E, 7997.250 N), D (2959.390 E, 7487.090 N).

Sides and angles of triangle PWD from the coordinates: $P = 47°14'34"$, $W = 69°55'56"$, $D = 62°49'30"$.

$K_P = 1.26664$, $K_W = 0.70075$, $K_D = 0.57227$ giving O = (3152.842 E, 8259.775 N).

#### Three solutions

| Set | $\angle POW$ | $\angle WOD$ | $\angle DOP$ | E (m) | N (m) |
|---|---|---|---|---|---|
| 1 | 140°58'51" | 82°18'14" | 136°42'55" | 3152.842 | 8259.775 |
| 2 | 140°58'21" | 82°18'14" | 136°43'25" | 3152.793 | 8259.754 |
| 3 | 140°58'51" | 82°17'44" | 136°43'25" | 3152.777 | 8259.837 |

#### Average

$$
E_O = \frac{3152.842 + 3152.793 + 3152.777}{3} = 3152.804\ \text{m},\qquad N_O = \frac{8259.775 + 8259.754 + 8259.837}{3} = 8259.788\ \text{m}
$$

**Answer: average coordinates of O = (3152.804 E, 8259.788 N).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With readings P = 90°00'00" and W = 230°58'51", what is ∠POW?
   A. 140°58'51"
   B. 230°58'51"
   C. 320°58'51"
   D. 39°01'09"
   Answer: A. The angle is the difference of the readings: 230°58'51" - 90°00'00" = 140°58'51". (Key point: Angles from horizontal circle readings)

2. Why are three solutions computed?
   A. Three different methods are compared
   B. The closing reading of P differs by 30" so the three angles do not close; each pair of angles (third = 360° minus sum) gives a solution
   C. There are three instrument stations
   D. The three known points are re-numbered each time
   Answer: B. The first and closing readings on P differ by 30", so the third angle can be taken in two ways, giving three sets of angle pairs each solved by Tienstra. (Key point: Why there are three solutions)

3. How are the final coordinates of O obtained?
   A. The set with the smallest correction is chosen
   B. The coordinates from set 1 only are used
   C. The three solutions are averaged: O = (3152.804 E, 8259.788 N)
   D. The sum of the three sets is used
   Answer: C. The mean of the three Tienstra solutions gives the average coordinates 3152.804 E, 8259.788 N. (Key point: Final coordinates)

## Related in Surveying II

- [Coordinates of instrument station O and station A](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-pwd-instrument-station-a) — Surveying II, chapter 5
- [Derive three point analytical resection](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Athree-point-analytical-resection-derivation) — Surveying II, chapter 5, asked 2×
- [Resection point O from stations A, B, C](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-abc-2075b) — Surveying II, chapter 5
- [Resection point O inside triangle N, P, G](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-nag-triangle-2078c) — Surveying II, chapter 5
- [Analytical resection, derive coordinate expression](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanalytical-resection-derive-three-known) — Surveying II, chapter 5
- [Resection, two point problem, intersection](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-two-point-problem-ab-p) — Surveying II, chapter 5

## Related in other subjects

- [Mean angles and VCR, station O (A-D)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-vcr-120-15-10-ls10) — Surveying I, chapter 7, asked 3×
- [Mean angles and VCR, station O (reading 179°59′40″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-first-reversed-179-59-40) — Surveying I, chapter 7, asked 3×
- [Resection by three point problem](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Athree-point-resection-sketches) — Surveying I, chapter 6
- [Mean horizontal angles, station O (P, Q, R)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-o-pqrp) — Surveying I, chapter 7