---
title: "Tienstra's method, coordinates of O"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:tienstra-pqr-observation-o"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/tienstra-pqr-observation-o
site: BCT Engineering
---

# Tienstra's method, coordinates of O

Surveying II · Chapter 5: Orientation

- Asked in: 2078 Poush, 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/tienstra-pqr-observation-o
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

The coordinates of three stations P, Q and R are given in the table and from an instrument point O following observations are taken.

| Stations | Northing (m) | Easting (m) | Angle to right |
|---|---|---|---|
| P | 5000.00 | 5000.00 | ∠POQ = 114°30'18" |
| Q | 9000.00 | 8500.00 | ∠QOR = 122°20'32" |
| R | 5000.00 | 12000.00 | ∠ROP = 123°09'10" |

Calculate the coordinates of 'O' by Tienstra's method.

## Short answer

Compute triangle PQR interior angles from given coordinates. Tienstra constants K_i = 1/(cot A_i - cot α_i) yield weighted mean coordinates: O = 8615.199 m E, 6892.720 m N.

## Answer

Coordinates are written as (Easting, Northing); the table gives Northing first, so P = (5000, 5000), Q = (8500 E, 9000 N), R = (12000 E, 5000 N). Observed: $\angle POQ = 114^\circ30'18''$, $\angle QOR = 122^\circ20'32''$, $\angle ROP = 123^\circ09'10''$ (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) |
|---|---|---|
| P | 5000.000 | 5000.000 |
| Q | 8500.000 | 9000.000 |
| R | 12000.000 | 5000.000 |

Angles at O: $\angle POQ = \gamma = 114°30'18"$, $\angle QOR = \alpha = 122°20'32"$, $\angle ROP = \beta = 123°09'10"$ (sum = 360°00'00").

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

| Side | Bearing | Length (m) |
|---|---|---|
| PQ | 41°11'09" | 5315.073 |
| QR | 138°48'51" | 5315.073 |
| RP | 270°00'00" | 7000.000 |

Interior angles: $A$ (at P) $= 48°48'51"$, $B$ (at Q) $= 82°22'19"$, $C$ (at R) $= 48°48'51"$ (sum 180°00'00").

#### Step 2: Tienstra constants

$$
\begin{aligned}
K_1 &= \frac{1}{\cot 48°48' - \cot 122°20'32"} = 0.66304 \\
K_2 &= \frac{1}{\cot 82°22' - \cot 123°09'10"} = 1.27043 \\
K_3 &= \frac{1}{\cot 48°48' - \cot 114°30'18"} = 0.75141
\end{aligned}
$$

$\Sigma K = 2.68488$

#### Step 3: Coordinates of O

$$
E_O = \frac{(0.66304)(5000.000) + (1.27043)(8500.000) + (0.75141)(12000.000)}{2.68488} = 8615.199\ \text{m}
$$

$$
N_O = \frac{(0.66304)(5000.000) + (1.27043)(9000.000) + (0.75141)(5000.000)}{2.68488} = 6892.720\ \text{m}
$$

**Check:** the angles computed back from these coordinates are 114°30'18", 122°20'32" and 123°09'10", equal to the observed angles.

**Answer: coordinates of O = (8615.199 E, 6892.720 N).**

## Check yourself

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

1. The table lists Northing first, then Easting. Which is the correct coordinate of Q (Northing 9000, Easting 8500)?
   A. (9000 E, 8500 N)
   B. (8500 E, 9000 N)
   C. (8500 E, 8500 N)
   D. (9000 E, 9000 N)
   Answer: B. Coordinates are used as (Easting, Northing), so Q = (8500 E, 9000 N). Swapping them would give a wrong triangle. (Key point: Reading the coordinate table)

2. Do the observed angles need a misclosure correction here?
   A. Yes, 30" must be removed
   B. Yes, 10" must be added
   C. No, 114°30'18" + 122°20'32" + 123°09'10" = 360° exactly
   D. No, because only two angles are needed
   Answer: C. The three angles add to exactly 360°00'00", so they are used as observed. (Key point: Angle sum check)

3. Which observed angle goes with the constant K1 for station P (interior angle A at P)?
   A. Angle POQ
   B. Angle ROP
   C. Angle QOR, subtended by the opposite side QR
   D. Angle PQR
   Answer: C. K1 uses A together with the angle at O subtended by the side opposite P, namely QR, so alpha = angle QOR = 122°20'32". (Key point: Matching angles to stations)

4. With K = 0.66304, 1.27043, 0.75141 for P, Q, R, the Easting of O is found as:
   A. (sum of K x E)/(sum of K) = 8615.199 m
   B. The mean of 5000, 8500 and 12000
   C. The sum of K x E without dividing
   D. The Easting of Q alone
   Answer: A. E_O is the K-weighted mean of the station Eastings, divided by the sum of K (2.68488), giving 8615.199 m. (Key point: Final coordinates by weighted mean)

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