---
title: "Resection point O inside triangle N, P, G"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:resection-nag-triangle-2078c"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-nag-triangle-2078c
site: BCT Engineering
---

# Resection point O inside triangle N, P, G

Surveying II · Chapter 5: Orientation

- Asked in: 2078 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-nag-triangle-2078c
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

The co-ordinates of three stations are known Nagarkot (N), Phulchowki (P) and Nagarjun (G). A resection point "O" is set inside the triangle and the observations taken for horizontal angle to these known stations from "O" are given in the table below along with the co-ordinates of the three known stations. Calculate the co-ordinate of resection point "O".

| Known station | Easting (m) | Northing (m) | Horizontal Angle |
|---|---|---|---|
| Nagarkot (N) | 352836.105 | 3066097.505 | ∠NOP = 140°04'45" |
| Phulchowki (P) | 342615.318 | 3050525.416 | ∠POG = 99°25'48" |
| Nagarjun (G) | 329189.716 | 3070164.918 | ∠GON = 120°28'57" |

## Short answer

Applying Tienstra's method with adjusted angles ∠NOP=140°04'55", ∠POG=99°25'58", ∠GON=120°29'07", and triangle angles from the given N,P,G coordinates, the unknown point O has coordinates (346069.994 mE, 3061715.988 mN).

## Answer

The observed angles are $\angle NOP = 140^\circ04'45''$, $\angle POG = 99^\circ25'48''$, $\angle GON = 120^\circ28'57''$. Their sum is 359°59'30", a misclosure of 30". It is shared equally (+10" to each angle):
$\angle NOP = 140^\circ04'55''$, $\angle POG = 99^\circ25'58''$, $\angle GON = 120^\circ29'07''$.

**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 the unknown point 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) |
|---|---|---|
| N | 352836.105 | 3066097.505 |
| P | 342615.318 | 3050525.416 |
| G | 329189.716 | 3070164.918 |

Angles at O: $\angle NOP = \gamma = 140°04'55"$, $\angle POG = \alpha = 99°25'58"$, $\angle GON = \beta = 120°29'07"$ (sum = 360°00'00").

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

| Side | Bearing | Length (m) |
|---|---|---|
| NP | 213°16'44" | 18626.713 |
| PG | 325°38'36" | 23789.847 |
| GN | 99°45'36" | 23993.657 |

Interior angles: $A$ (at N) $= 66°28'51"$, $B$ (at P) $= 67°38'08"$, $C$ (at G) $= 45°53'01"$ (sum 180°00'00").

#### Step 2: Tienstra constants

$$
\begin{aligned}
K_1 &= \frac{1}{\cot 66°28'51" - \cot 99°25'58"} = 1.66294 \\
K_2 &= \frac{1}{\cot 67°38'08" - \cot 120°29'07"} = 0.99986 \\
K_3 &= \frac{1}{\cot 45°53'01" - \cot 140°04'55"} = 0.46193
\end{aligned}
$$

$\Sigma K = 3.12472$

#### Step 3: Coordinates of O

$$
E_O = \frac{(1.66294)(352836.105) + (0.99986)(342615.318) + (0.46193)(329189.716)}{3.12472} = 346069.994\ \text{m}
$$

$$
N_O = \frac{(1.66294)(3066097.505) + (0.99986)(3050525.416) + (0.46193)(3070164.918)}{3.12472} = 3061715.988\ \text{m}
$$

**Check:** the angles computed back from these coordinates are 140°04'55", 99°25'58" and 120°29'07", equal to the adjusted observed angles.

**Answer: coordinates of O = (346069.994 E, 3061715.988 N).**

## Check yourself

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

1. The observed angles at O sum to 359°59'30". How is this corrected?
   A. Subtract 30" from one angle
   B. Add 10" to each of the three angles
   C. Add 30" to each angle
   D. Discard the observations
   Answer: B. The total must be 360°, so the 30" shortfall is shared equally: +10" to each angle. (Key point: Misclosure correction)

2. How are the interior angles A, B, C of triangle NPG obtained?
   A. Measured with the theodolite at O
   B. Computed from bearings of the sides found from the coordinates
   C. Taken equal to the angles at O
   D. Assumed to be 60° each
   Answer: B. Bearings and lengths of NP, PG and GN come from the coordinates, and the differences of bearings give the interior angles, which sum to 180°. (Key point: Interior angles of the triangle)

3. For the constant K1 at Nagarkot (interior angle A = 66°28'51"), which observed angle is used?
   A. Angle NOP = 140°04'55"
   B. Angle GON = 120°29'07"
   C. Angle POG = 99°25'58", subtended by the opposite side PG
   D. Angle NPG = 67°38'08"
   Answer: C. K1 = 1/(cot A - cot alpha) where alpha is the angle at O subtended by the side opposite N, i.e. angle POG. (Key point: Constant for station N)

4. How does the answer verify the computed position of O?
   A. Angles recomputed from the coordinates of O equal the observed (corrected) angles
   B. The sum of the K values equals 1
   C. The coordinates equal those of the centroid
   D. The computed Easting equals the Northing
   Answer: A. Back-computing the angles at O from the final coordinates reproduces the observed angles, confirming the solution. (Key point: Verification of result)

## Related in Surveying II

- [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
- [Coordinates of O from Swoyambhu, Harisiddhi, Chovar](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-swoyambhu-harisiddhi-chovar) — Surveying II, chapter 5
- [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
- [Intersection vs resection, three point method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aintersection-vs-resection-three-point) — 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
- [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×

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