---
title: "Coordinates of O from Swoyambhu, Harisiddhi, Chovar"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:resection-swoyambhu-harisiddhi-chovar"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-swoyambhu-harisiddhi-chovar
site: BCT Engineering
---

# Coordinates of O from Swoyambhu, Harisiddhi, Chovar

Surveying II · Chapter 5: Orientation

- Asked in: 2074 Bhadra, 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-swoyambhu-harisiddhi-chovar
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

The co-ordinates of three known stations Swoyambhu (S), Harisiddhi (D) and Chovar Temple (C) are given below. A traverse point "O" is set outside of this triangle and observations are taken for horizontal angle to these known co-ordinate points. Calculate the co-ordinate of station point "O".

| Known Station | Horizontal Angle | Easting, m | Northing, m |
|---|---|---|---|
| Swoyambhu (S) | ∠SOD = 40°20'21" | 627465.718 | 3066929.474 |
| Harisiddhi (D) | ∠DOC = 98°56'41" | 629603.054 | 6065364.275 |
| Chovar Temple (C) | ∠COS = 220°42'58" | 627612.753 | 3061479.468 |

## Short answer

Correct Harisiddhi northing to 3065364.275 m, compute triangle SDC interior angles, then Tienstra constants. O = 626714.801 m E, 3063049.803 m N.

## Answer

The northing of Harisiddhi is printed as 6065364.275; it is taken as 3065364.275 (a misprint, since the other stations lie near 3 06x xxx N). O lies outside the triangle SDC. $\angle SOD = 40^\circ20'21''$, $\angle DOC = 98^\circ56'41''$, $\angle COS = 220^\circ42'58''$ (sum 360°).

**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) |
|---|---|---|
| S | 627465.718 | 3066929.474 |
| D | 629603.054 | 3065364.275 |
| C | 627612.753 | 3061479.468 |

Angles at O: $\angle SOD = \gamma = 40°20'21"$, $\angle DOC = \alpha = 98°56'41"$, $\angle COS = \beta = 220°42'58"$ (sum = 360°00'00").

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

| Side | Bearing | Length (m) |
|---|---|---|
| SD | 126°12'57" | 2649.161 |
| DC | 207°07'39" | 4364.977 |
| CS | 358°27'17" | 5451.989 |

Interior angles: $A$ (at S) $= 52°14'20"$, $B$ (at D) $= 99°05'18"$, $C$ (at C) $= 28°40'22"$ (sum 180°00'00").

#### Step 2: Tienstra constants

$$
\begin{aligned}
K_1 &= \frac{1}{\cot 52°14' - \cot 98°56'41"} = 1.07297 \\
K_2 &= \frac{1}{\cot 99°05' - \cot 220°42'58"} = -0.75648 \\
K_3 &= \frac{1}{\cot 28°40' - \cot 40°20'21"} = 1.53593
\end{aligned}
$$

$\Sigma K = 1.85242$

#### Step 3: Coordinates of O

$$
E_O = \frac{(1.07297)(627465.718) + (-0.75648)(629603.054) + (1.53593)(627612.753)}{1.85242} = 626714.801\ \text{m}
$$

$$
N_O = \frac{(1.07297)(3066929.474) + (-0.75648)(3065364.275) + (1.53593)(3061479.468)}{1.85242} = 3063049.803\ \text{m}
$$

**Check:** the angles computed back from these coordinates are 40°20'21", 98°56'41" and 220°42'58", equal to the observed angles.

**Answer: coordinates of O = (626714.801 E, 3063049.803 N).**

## Check yourself

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

1. Harisiddhi's northing is printed as 6065364.275 while the others are near 3 06x xxx. What is done?
   A. It is used as printed
   B. It is taken as 3065364.275 as a misprint, and the answer states this
   C. Harisiddhi is dropped from the problem
   D. The other stations are changed to 6 06x xxx
   Answer: B. The value is clearly a typing error, so it is read as 3065364.275 and the assumption is stated. (Key point: Handling a printed misprint)

2. The angle COS = 220°42'58" shows that O is:
   A. Inside the triangle SDC
   B. Outside the triangle SDC
   C. On the line SD
   D. At station C
   Answer: B. For a point inside the triangle each angle is less than 180°; an angle above 180° means O lies outside. (Key point: Observed angle above 180°)

3. K3 uses the interior angle at C with which angle at O?
   A. Angle DOC = 98°56'41"
   B. Angle COS = 220°42'58"
   C. Angle SOD = 40°20'21", subtended by the opposite side SD
   D. Angle at S
   Answer: C. The side opposite C is SD, so the angle at O is SOD = 40°20'21". (Key point: Constant K3 inputs)

4. K2 = -0.75648 for D. How does it enter the result?
   A. Its absolute value is used
   B. It is used with its negative sign in sum(KE)/sum(K)
   C. It is set to zero
   D. It is added to the Northing
   Answer: B. Negative weights are valid for a point outside the triangle and are kept with their sign, giving sum K = 1.85242. (Key point: Sign of K and weighted mean)

## Related in Surveying II

- [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
- [Tienstra's method, X from Swayambhu, Dharara, Chovar](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atienstra-swayambhu-dharara-chovar) — Surveying II, chapter 5
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- [Coordinates of X from S, D, C](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-sdc-theodolite-2077c) — 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
- [RL of temple top from stations A and B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atemple-top-rl-two-stations) — Surveying II, chapter 3

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