---
title: "Distances OA and OC by resection on A, B, C"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:resection-abc-sides-oa-oc"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-abc-sides-oa-oc
site: BCT Engineering
---

# Distances OA and OC by resection on A, B, C

Surveying II · Chapter 5: Orientation

- Asked in: 2070 Magh, 10 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-abc-sides-oa-oc
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

A, B and C are three visible stations in a location survey. The computed sides of triangle ABC are AB = 1200 m [?], BC = 1442 m and CA = 1960 m. A station 'O' is established outside the triangle and its position is to be determined by resection on A, B and C. The angles AOB and BOC being 45°30' and 52°15' respectively. Determine the distances of OA and OC.

## Short answer

Using the tangent method for resection, the unknown angles at A and C are solved from quadrilateral OABC. Then OA = 884.01 m and OC = 1634.17 m.

## Answer

The sides AB = 1200 m, BC = 1442 m and CA = 1960 m are used as given (the value marked [?] in the question is taken as printed). The angles $\alpha = \angle AOB = 45^\circ30'$ and $\beta = \angle BOC = 52^\circ15'$ are observed at O. B is the middle station. The method is the tangent method (Pothenot's problem).

#### Step 1: Angle B of triangle ABC

$$
\cos B = \frac{AB^2 + BC^2 - CA^2}{2\,AB\cdot BC} = \frac{1200^2 + 1442^2 - 1960^2}{2(1200)(1442)} = -0.09311
$$

$B = 95°20'33"$

#### Step 2: Sum and ratio of the unknown angles

Let $x = \angle OAB$ and $y = \angle OCB$. In the quadrilateral OABC:

$$
x + y = S = 360^\circ - (\alpha + \beta + B) = 360^\circ - (45^\circ30' + 52^\circ15' + 95°20'33") = 166°54'27"
$$

$$
k = \frac{\sin x}{\sin y} = \frac{BC\sin\alpha}{AB\sin\beta} = \frac{1442\sin 45^\circ30'}{1200\sin 52^\circ15'} = 1.08398
$$

#### Step 3: Solve for x and y

$$
\tan\frac{x-y}{2} = \frac{k-1}{k+1}\tan\frac{S}{2} = \frac{0.08398}{2.08398}\tan 83°27'13" \quad\Rightarrow\quad \frac{x-y}{2} = 19°20'56"
$$

$$
x = \frac{S}{2} + 19°20'56" = 102°48'10",\qquad y = \frac{S}{2} - 19°20'56" = 64°06'17"
$$

#### Step 4: Distances OA and OC

In triangle AOB: $\angle ABO = 180^\circ - \alpha - x = 31°41'50"$

$$
OA = \frac{AB\sin\angle ABO}{\sin\alpha} = \frac{1200\sin 31°41'50"}{\sin 45^\circ30'} = 884.005\ \text{m}
$$

In triangle BOC: $\angle CBO = 180^\circ - \beta - y = 63°38'43"$

$$
OC = \frac{BC\sin\angle CBO}{\sin\beta} = \frac{1442\sin 63°38'43"}{\sin 52^\circ15'} = 1634.172\ \text{m}
$$

**Check:** OB from triangle AOB $= \frac{1200\sin x}{\sin\alpha} = 1640.611$ m and from triangle BOC $= \frac{1442\sin y}{\sin\beta} = 1640.611$ m. They agree.

**Answer: OA = 884.01 m and OC = 1634.17 m (OB = 1640.61 m).**

## Check yourself

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

1. cos B = (AB^2 + BC^2 - CA^2)/(2 AB BC) comes out negative (-0.09311). What does this mean?
   A. Angle B is acute
   B. Angle B is obtuse (95°20'33")
   C. Angle B is zero
   D. The triangle does not exist
   Answer: B. A negative cosine means the angle exceeds 90°; here B = 95°20'33". (Key point: Angle B by cosine rule)

2. With alpha = 45°30', beta = 52°15' and B = 95°20'33", S = x + y is:
   A. 166°54'27"
   B. 193°05'33"
   C. 180°
   D. 93°27'13"
   Answer: A. S = 360° - (45°30' + 52°15' + 95°20'33") = 166°54'27". (Key point: Sum of x and y)

3. How is k = sin x / sin y obtained?
   A. k = AB sin alpha / (BC sin beta)
   B. k = BC sin alpha / (AB sin beta) = 1.08398
   C. k = CA / AB
   D. k = sin alpha / sin beta
   Answer: B. Equating OB from triangles AOB and BOC gives sin x / sin y = BC sin alpha / (AB sin beta). (Key point: Ratio k)

4. OA is found from triangle AOB as:
   A. AB sin(ABO) / sin(alpha), with ABO = 180° - alpha - x
   B. AB sin x / sin beta
   C. AB cos alpha
   D. BC sin y / sin alpha
   Answer: A. OA is opposite angle ABO and AB is opposite alpha, giving OA = 1200 sin 31°41'50" / sin 45°30' = 884.005 m. (Key point: Distance OA)

## 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
- [Resection point O from stations A, B, C](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-abc-2075b) — 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 hill station P from stations R and Q](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-signal-r-q-p) — Surveying II, chapter 3
- [Derive three point analytical resection](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Athree-point-analytical-resection-derivation) — Surveying II, chapter 5, asked 2×
- [Intersection vs resection, three point method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aintersection-vs-resection-three-point) — Surveying II, chapter 5

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