---
title: "Coordinates of X from S, D, C"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:resection-sdc-theodolite-2077c"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-sdc-theodolite-2077c
site: BCT Engineering
---

# Coordinates of X from S, D, C

Surveying II · Chapter 5: Orientation

- Asked in: 2077 Chaitra, 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-sdc-theodolite-2077c
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

The coordinates of three known stations (S), (D) and (C) are given in Table below. A theodolite is setup over the unknown point X, which is set outside of this known triangle and horizontal angle observation are made. Calculate coordinates of station X using any one method.

| Known Points | Horizontal angle | Easting (m) | Northing (m) |
|---|---|---|---|
| (S) | ∠SXD = 41°20'21" | 7464.718 | 6928.474 |
| (D) | ∠DXC = 97°56'41" | 9602.054 | 5363.275 |
| (C) | ∠CXS = 220°42'48" | 7611.753 | 1479.468 |

## Short answer

Adjust observed angles for 10" misclosure, compute triangle SDC interior angles, then Tienstra's constants. Result: X = 6677.417 m E, 3174.042 m N.

## Answer

X lies outside the triangle SDC. The angles $\angle SXD = 41^\circ20'21''$, $\angle DXC = 97^\circ56'41''$, $\angle CXS = 220^\circ42'48''$ sum to 359°59'50" (misclosure 10"), which is shared equally (+3.3" each): 41°20'24", 97°56'44", 220°42'51". Angles above 180° are used as measured (clockwise); the formula holds for a station outside the triangle.

**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 | 7464.718 | 6928.474 |
| D | 9602.054 | 5363.275 |
| C | 7611.753 | 1479.468 |

Angles at X: $\angle SXD = \gamma = 41°20'24"$, $\angle DXC = \alpha = 97°56'44"$, $\angle CXS = \beta = 220°42'51"$ (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

$$
\begin{aligned}
K_1 &= \frac{1}{\cot 52°14' - \cot 97°56'44"} = 1.09388 \\
K_2 &= \frac{1}{\cot 99°04' - \cot 220°42'51"} = -0.75650 \\
K_3 &= \frac{1}{\cot 28°40' - \cot 41°20'24"} = 1.44623
\end{aligned}
$$

$\Sigma K = 1.78362$

#### Step 3: Coordinates of X

$$
E_X = \frac{(1.09388)(7464.718) + (-0.75650)(9602.054) + (1.44623)(7611.753)}{1.78362} = 6677.417\ \text{m}
$$

$$
N_X = \frac{(1.09388)(6928.474) + (-0.75650)(5363.275) + (1.44623)(1479.468)}{1.78362} = 3174.042\ \text{m}
$$

**Check:** the angles computed back from these coordinates are 41°20'24", 97°56'44" and 220°42'51", equal to the observed angles.

**Answer: coordinates of X = (6677.417 E, 3174.042 N).**

## Check yourself

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

1. X lies outside triangle SDC and one observed angle is 220°42'48". How is it treated?
   A. It is replaced by 360° minus the angle
   B. It is used as measured (clockwise), because the formula also holds outside the triangle
   C. The problem cannot be solved
   D. It is halved
   Answer: B. Angles above 180° are used as measured clockwise; Tienstra's formula remains valid for a station outside the triangle. (Key point: Station outside the triangle)

2. The angles sum to 359°59'50". What correction is applied to each angle?
   A. -10" each
   B. +10" to one angle only
   C. About +3.3" each
   D. +30" each
   Answer: C. The 10" shortfall is shared equally among three angles, about +3.3" each. (Key point: Correction of misclosure)

3. K2 comes out negative (-0.7565). What does this mean?
   A. An arithmetic mistake that must be redone
   B. It is a valid weight; negative weights occur when X is outside the triangle, and the sum of K is still used
   C. The station D must be dropped
   D. X lies on the danger circle
   Answer: B. For a point outside the triangle one or more K values can be negative. They are used with their signs in the weighted mean. (Key point: Negative K value)

4. K2 is computed from which pair of angles?
   A. Interior angle B at D and the angle CXS (220°42'51")
   B. Interior angle A at S and the angle DXC
   C. Interior angle C and the angle SXD
   D. Angle B and the angle DXC
   Answer: A. K2 pairs the triangle angle at D with the angle at X subtended by the opposite side CS, which is angle CXS. (Key point: Constant K2 inputs)

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