---
title: "Tienstra's method, O from P, Q, R"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:tienstra-pqr-observed-angles-2081c"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/tienstra-pqr-observed-angles-2081c
site: BCT Engineering
---

# Tienstra's method, O from P, Q, R

Surveying II · Chapter 5: Orientation

- Asked in: 2081 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/tienstra-pqr-observed-angles-2081c
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

The coordinates of three known stations P, Q and R are given below and from an instrument point O following observations were taken. Calculate the coordinates of 'O' by Tienstra's method.

| Known station | Observed Angles | Northing (m) | Easting (m) |
|---|---|---|---|
| P | ∠POQ = 142°43'32" | 29236.48 | 24078.31 |
| Q | ∠ROQ = 92°19'52" | 31493.20 | 26266.48 |
| R | ∠POR = 124°56'26" | 29661.04 | 28377.67 |

## Short answer

Using Tienstra's method, interior angles of triangle PQR are computed from coordinates, then Tienstra constants from observed angles. Weighted average gives O = (25962.348 m E, 30549.246 m N).

## Answer

Coordinates are given as Northing, Easting; they are used here as (E, N): P = (24078.31 E, 29236.48 N), Q = (26266.48 E, 31493.20 N), R = (28377.67 E, 29661.04 N).

Observed: $\angle POQ = 142^\circ43'32''$, $\angle QOR = \angle ROQ = 92^\circ19'52''$, $\angle ROP = \angle POR = 124^\circ56'26''$. The sum is 359°59'50" (misclosure 10"), shared equally (+3.3" to each): 142°43'35", 92°19'55", 124°56'29".

**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 | 24078.310 | 29236.480 |
| Q | 26266.480 | 31493.200 |
| R | 28377.670 | 29661.040 |

Angles at O: $\angle POQ = \gamma = 142°43'35"$, $\angle QOR = \alpha = 92°19'55"$, $\angle ROP = \beta = 124°56'29"$ (sum = 360°00'00").

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

| Side | Bearing | Length (m) |
|---|---|---|
| PQ | 44°06'59" | 3143.386 |
| QR | 130°57'09" | 2795.341 |
| RP | 264°21'37" | 4320.272 |

Interior angles: $A$ (at P) $= 40°14'38"$, $B$ (at Q) $= 93°09'50"$, $C$ (at R) $= 46°35'32"$ (sum 180°00'00").

#### Step 2: Tienstra constants

$$
\begin{aligned}
K_1 &= \frac{1}{\cot 40°14' - \cot 92°19'55"} = 0.81818 \\
K_2 &= \frac{1}{\cot 93°09' - \cot 124°56'29"} = 1.55422 \\
K_3 &= \frac{1}{\cot 46°35' - \cot 142°43'35"} = 0.44251
\end{aligned}
$$

$\Sigma K = 2.81490$

#### Step 3: Coordinates of O

$$
E_O = \frac{(0.81818)(24078.310) + (1.55422)(26266.480) + (0.44251)(28377.670)}{2.81490} = 25962.348\ \text{m}
$$

$$
N_O = \frac{(0.81818)(29236.480) + (1.55422)(31493.200) + (0.44251)(29661.040)}{2.81490} = 30549.246\ \text{m}
$$

**Check:** the angles computed back from these coordinates are 142°43'35", 92°19'55" and 124°56'29", equal to the observed angles.

**Answer: coordinates of O = (25962.348 E, 30549.246 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°59'50". What is done before using Tienstra's formula?
   A. The 10" misclosure is shared equally (about 3.3" to each angle) so the sum is 360°
   B. The 10" is added to the largest angle only
   C. The 10" is ignored and 360° is not required
   D. The smallest angle is recomputed as 360° minus the other two
   Answer: A. The angles at the unknown point must total 360°, so the small misclosure is distributed equally among the three angles. (Key point: Adjusting the angles at O)

2. In Tienstra's formula, how is K1 formed for known station S1?
   A. K1 = 1 / (cot A + cot α)
   B. K1 = 1 / (cot A - cot α), with A the triangle angle at S1 and α the observed angle opposite S1 (S2-P-S3)
   C. K1 = 1 / (tan A - tan α)
   D. K1 = cot A - cot α
   Answer: B. Each K is the reciprocal of (cot of the triangle angle at the station minus cot of the observed angle subtended by the opposite side). (Key point: Tienstra constant)

3. How are the coordinates of O finally obtained?
   A. As the simple average of the three known stations
   B. As E = ΣK·E / ΣK and N = ΣK·N / ΣK, i.e. a weighted mean using K as weights
   C. As the coordinates of the station with the largest K
   D. By adding K to the coordinates of the nearest station
   Answer: B. E_O = (K1E1+K2E2+K3E3)/(K1+K2+K3) and likewise for N, giving O = (25962.348 E, 30549.246 N). (Key point: Weighted mean for coordinates)

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