---
title: "Tienstra's method, X from Swayambhu, Dharara, Chovar"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:tienstra-swayambhu-dharara-chovar"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/tienstra-swayambhu-dharara-chovar
site: BCT Engineering
---

# Tienstra's method, X from Swayambhu, Dharara, Chovar

Surveying II · Chapter 5: Orientation

- Asked in: 2071 Bhadra, 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/tienstra-swayambhu-dharara-chovar
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

The co-ordinates of three known stations Swayambhu, Dharara and Chovar temple is given below.

| Known station | Horizontal angle | Easting | Northing |
|---|---|---|---|
| Swayambhu (S) | ∠SXD = 41°20'21" | 627464.718 | 3066928.474 |
| Dharara (D) | ∠DXC = 97°56'41" | 629602.054 | 3065363.275 |
| Chovar Temple (C) | ∠CXS = 220°42'58" | 627611.753 | 3061479.468 |

A theodolite is set up over an unknown point X. Calculate the co-ordinates of station X using the Tienstra method.

## Short answer

Triangle SDC interior angles: S=52°14'19", D=99°04'57", C=28°40'45". Tienstra constants: K₁=1.09390, K₂=−0.75654, K₃=1.44631. Weighted average gives X = (626677.395 mE, 3063173.956 mN).

## Answer

X lies outside the triangle SDC: $\angle SXD = 41^\circ20'21''$, $\angle DXC = 97^\circ56'41''$, $\angle CXS = 220^\circ42'58''$ (sum 360°). The angle above 180° is used as measured clockwise.

**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 | 627464.718 | 3066928.474 |
| D | 629602.054 | 3065363.275 |
| C | 627611.753 | 3061479.468 |

Angles at X: $\angle SXD = \gamma = 41°20'21"$, $\angle DXC = \alpha = 97°56'41"$, $\angle CXS = \beta = 220°42'58"$ (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°08'00" | 4364.087 |
| CS | 358°27'16" | 5450.989 |

Interior angles: $A$ (at S) $= 52°14'19"$, $B$ (at D) $= 99°04'57"$, $C$ (at C) $= 28°40'45"$ (sum 180°00'00").

#### Step 2: Tienstra constants

$$
\begin{aligned}
K_1 &= \frac{1}{\cot 52°14' - \cot 97°56'41"} = 1.09390 \\
K_2 &= \frac{1}{\cot 99°04' - \cot 220°42'58"} = -0.75654 \\
K_3 &= \frac{1}{\cot 28°40' - \cot 41°20'21"} = 1.44631
\end{aligned}
$$

$\Sigma K = 1.78367$

#### Step 3: Coordinates of X

$$
E_X = \frac{(1.09390)(627464.718) + (-0.75654)(629602.054) + (1.44631)(627611.753)}{1.78367} = 626677.395\ \text{m}
$$

$$
N_X = \frac{(1.09390)(3066928.474) + (-0.75654)(3065363.275) + (1.44631)(3061479.468)}{1.78367} = 3063173.956\ \text{m}
$$

**Check:** the angles computed back from these coordinates are 41°20'21", 97°56'41" and 220°42'58", equal to the observed angles.

**Answer: coordinates of X = (626677.395 E, 3063173.956 N).**

## Check yourself

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

1. The angle CXS = 220°42'58" exceeds 180°. What does this indicate and how is it used?
   A. X is outside the triangle; it is used as measured clockwise
   B. X is inside the triangle; it is replaced by 139°17'02"
   C. The observation is wrong
   D. X is at station C
   Answer: A. An angle above 180° shows that X lies outside the triangle; it is used as measured. (Key point: Reflex angle meaning)

2. K1 (at S, A = 52°14'19") uses which observed angle?
   A. Angle SXD
   B. Angle CXS
   C. Angle DXC = 97°56'41", subtended by the opposite side DC
   D. Angle at D
   Answer: C. The side opposite S is DC, so alpha = angle DXC. (Key point: Pairing for K1)

3. K = 1.09390, -0.75654, 1.44631. The denominator in the coordinate formulas is:
   A. 1.78367, the algebraic sum including the negative term
   B. 3.29675, the sum of absolute values
   C. 1
   D. 1.44631
   Answer: A. The negative K is kept, so the sum is 1.09390 - 0.75654 + 1.44631 = 1.78367. (Key point: Sum of K)

4. How is the final X = (626677.395 E, 3063173.956 N) checked?
   A. Angles computed back from X equal the observed angles
   B. The K values add to 1
   C. The Northing exceeds the Easting
   D. Distances equal the sides of the triangle
   Answer: A. Recomputing angles SXD, DXC and CXS from the coordinates gives 41°20'21", 97°56'41" and 220°42'58". (Key point: Verification)

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