---
title: "Resection point O from stations A, B, C"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:resection-abc-2075b"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-abc-2075b
site: BCT Engineering
---

# Resection point O from stations A, B, C

Surveying II · Chapter 5: Orientation

- Asked in: 2075 Baisakh, 7 marks
- Video: [Blunt's Method for Resection // Surveying](https://www.youtube.com/watch?v=ZLueu5FMl7k) — engiNERDing
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-abc-2075b
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

Stations A, B and C have the following respective coordinates (2876.24 mE, 8754.11 mN), (3810.80 mE, 7997.25 mN) and (2959.39 mE, 7487.09 mN) respectively. Station 'O' was established and following observations were recorded by a theodolite.

| Pointing towards | Horizontal Circle Readings |
|---|---|
| A | 90°00'00" |
| B | 230°58'51" |
| C | 313°17'05" |
| A | 90°00'30" |

Determine the coordinates of resection point 'O'.

## Short answer

Correct 30" closure in observed angles, then use Tienstra's method with triangle ABC interior angles. O = 3152.804 m E, 8259.788 m N.

## Answer

The three-point problem is solved by Tienstra's method.

#### Observed angles at O

Readings: A = 90°00'00", B = 230°58'51", C = 313°17'05", A (closing) = 90°00'30".

- $\angle AOB = 140^\circ58'51''$, $\angle BOC = 82^\circ18'14''$, $\angle COA = 136^\circ43'25''$ (sum 360°00'30")
- The closing error of 30" is shared equally (−10" each): $\angle AOB = 140^\circ58'41''$, $\angle BOC = 82^\circ18'04''$, $\angle COA = 136^\circ43'15''$.

**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) |
|---|---|---|
| A | 2876.240 | 8754.110 |
| B | 3810.800 | 7997.250 |
| C | 2959.390 | 7487.090 |

Angles at O: $\angle AOB = \gamma = 140°58'41"$, $\angle BOC = \alpha = 82°18'04"$, $\angle COA = \beta = 136°43'15"$ (sum = 360°00'00").

#### Step 1: Angles of the triangle ABC from the coordinates

| Side | Bearing | Length (m) |
|---|---|---|
| AB | 129°00'09" | 1202.597 |
| BC | 239°04'13" | 992.553 |
| CA | 356°14'43" | 1269.745 |

Interior angles: $A$ (at A) $= 47°14'34"$, $B$ (at B) $= 69°55'56"$, $C$ (at C) $= 62°49'30"$ (sum 180°00'00").

#### Step 2: Tienstra constants

$$
\begin{aligned}
K_1 &= \frac{1}{\cot 47°14' - \cot 82°18'04"} = 1.26672 \\
K_2 &= \frac{1}{\cot 69°55' - \cot 136°43'15"} = 0.70065 \\
K_3 &= \frac{1}{\cot 62°49' - \cot 140°58'41"} = 0.57231
\end{aligned}
$$

$\Sigma K = 2.53967$

#### Step 3: Coordinates of O

$$
E_O = \frac{(1.26672)(2876.240) + (0.70065)(3810.800) + (0.57231)(2959.390)}{2.53967} = 3152.804\ \text{m}
$$

$$
N_O = \frac{(1.26672)(8754.110) + (0.70065)(7997.250) + (0.57231)(7487.090)}{2.53967} = 8259.788\ \text{m}
$$

**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 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. From the readings A = 90°00'00", B = 230°58'51" the angle AOB is:
   A. 230°58'51"
   B. 140°58'51", the difference of the readings
   C. 320°58'51"
   D. 90°00'00"
   Answer: B. A horizontal angle is the difference of the circle readings on the two targets: 230°58'51" - 90°00'00" = 140°58'51". (Key point: Angles from circle readings)

2. Why is A observed a second time (90°00'30") at the end?
   A. To get a second value of angle BOC
   B. To close the round and reveal a 30" error that is distributed equally
   C. To change the zero setting
   D. To fix the bearing of OA
   Answer: B. The closing reading differs by 30" from the first, showing the misclosure; the angles sum to 360°00'30" and 10" is deducted from each. (Key point: Use of the closing reading)

3. For station A (interior angle A = 47°14'34") which angle at O is used?
   A. Angle AOB
   B. Angle COA
   C. Angle BOC, opposite side BC
   D. None, K1 uses only the triangle
   Answer: C. The angle at O subtended by the side opposite A (that is BC) is alpha = angle BOC = 82°18'04". (Key point: Pairing for K1)

4. After finding K1, K2, K3, the Easting of O is:
   A. The sum of K1E1 + K2E2 + K3E3 only
   B. The sum of K x E divided by the sum of K, giving 3152.804 m
   C. The average of the three Eastings
   D. K1 x E1
   Answer: B. Tienstra's weighted mean divides by the sum of K (2.53967). (Key point: Computing the final position)

## Related in Surveying II

- [Three point analytical resection, average coordinates of O](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Athree-point-resection-p-w-d) — Surveying II, chapter 5
- [Resection point P from L, M, N](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-lmn-p-2079j) — 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
- [Two point resection, point C from A, B, D](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atwo-point-resection-acb-dca-adb) — Surveying II, chapter 5
- [Two point resection, point C (A, B at 7928.474 N)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atwo-point-resection-c-a-b) — 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

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