--- title: "Analytical resection, derive coordinate expression" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:analytical-resection-derive-three-known" source: https://www.bctengineer.com/q/semester_4/surveying_ii/analytical-resection-derive-three-known site: BCT Engineering --- # Analytical resection, derive coordinate expression Surveying II · Chapter 5: Orientation - Asked in: 2069 Bhadra, 8 marks - Video: [Analytical method of Solving three point problem | Collin's method | Surveying | Prashant YT | BE |](https://www.youtube.com/watch?v=i-Af3CXWY4w) — Prashant YT - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/analytical-resection-derive-three-known - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question What is analytical resection? Derive an expression to find the co-ordinates of unknown points by observations to three known points. ## Short answer Analytical resection is a method to compute the coordinates of an unknown point by measuring horizontal angles to three known points. Using the tangent method, the sum and ratio of unknown angles x and y are derived from the quadrilateral angle sum and sine rule, then distances and bearings give the coordinates. ## Answer **Resection** is the method of finding the coordinates of an unknown station P by measuring the horizontal angles at P to three (or two) stations of known coordinates. The instrument is set up only at the unknown point. **Analytical resection** is resection in which the coordinates of P are computed from the observed angles and the known coordinates (by trigonometry), instead of being found graphically (for example, with a plane table or tracing paper). #### Three-point resection (Pothenot's tangent method) **Problem:** A, B and C are known stations (B is the middle one). The instrument is at the unknown station P, and the angles $\alpha = \angle APB$ and $\beta = \angle BPC$ are observed. Find the coordinates of P. ```text B /|\ c / | \ a / | \ A | C \ | / \ | / \|/ P angles at P: alpha (APB), beta (BPC) ``` **Derivation** From the coordinates, find the sides $c = AB$, $a = BC$ and the angle $B = \angle ABC$ (from the bearings of BA and BC). Let $x = \angle PAB$ and $y = \angle PCB$ be the unknown angles. In the quadrilateral PABC the sum of the angles is $360^\circ$: $$ x + y + \alpha + \beta + B = 360^\circ \quad\Rightarrow\quad x + y = S = 360^\circ - (\alpha + \beta + B) $$ In triangle APB, by the sine rule, $PB = \dfrac{c\sin x}{\sin\alpha}$. In triangle BPC, $PB = \dfrac{a\sin y}{\sin\beta}$. Equating, $$ \frac{\sin x}{\sin y} = \frac{a\sin\alpha}{c\sin\beta} = k $$ Using $\dfrac{\sin x - \sin y}{\sin x + \sin y} = \dfrac{\tan\frac{x-y}{2}}{\tan\frac{x+y}{2}}$: $$ \tan\frac{x-y}{2} = \frac{k-1}{k+1}\tan\frac{S}{2} $$ So $\dfrac{x-y}{2}$ is found, and with $x + y = S$ we get $x$ and $y$ separately. **Coordinates of P** 1. In triangle APB: $\angle ABP = 180^\circ - \alpha - x$, so $AP = \dfrac{c\sin\angle ABP}{\sin\alpha}$ (or $BP = \dfrac{c\sin x}{\sin\alpha}$). 2. Bearing of $AP$ = bearing of $AB$ $\pm\, x$ (the sign depends on the side on which P lies). 3. $E_P = E_A + AP\sin(\text{bearing }AP)$, $\;N_P = N_A + AP\cos(\text{bearing }AP)$. 4. Check by computing P from C: $CP = \dfrac{a\sin\angle CBP}{\sin\beta}$, with the bearing of $CP$ = bearing of $CB \mp y$. Both must give the same coordinates. **Note:** If P lies on the circle through A, B and C (the "danger circle"), then $\alpha + \beta + B = 180^\circ$, $\tan\frac{S}{2}$ is infinite, and the problem has no unique solution. The point P should be chosen away from this circle. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the three-point resection derivation, with x = angle PAB and y = angle PCB, what does the angle sum of quadrilateral PABC give? A. x + y = 180° - (α + β) B. x + y = 360° - (α + β) C. x + y = 360° - (α + β + B) D. x + y = (α + β + B) - 180° Answer: C. The four angles of quadrilateral PABC add to 360°, so x + y + α + β + B = 360°, giving x + y = S = 360° - (α + β + B). (Key point: Angle sum of quadrilateral PABC) 2. How is the ratio sin x / sin y obtained? A. By writing PB from triangle APB and from triangle BPC with the sine rule and equating them B. By applying the cosine rule to triangle ABC C. By writing AC from triangles APC and ABC D. By taking the ratio of the observed angles α and β Answer: A. PB = c sin x / sin α = a sin y / sin β, so sin x / sin y = a sin α / (c sin β) = k. (Key point: Sine rule on common side PB) 3. Which relation gives (x - y)/2 once k = sin x / sin y and S = x + y are known? A. tan((x-y)/2) = (k+1)/(k-1) tan(S/2) B. tan((x-y)/2) = (k-1)/(k+1) tan(S/2) C. sin((x-y)/2) = (k-1)/(k+1) sin(S/2) D. tan((x-y)/2) = k tan(S/2) Answer: B. Using (sin x - sin y)/(sin x + sin y) = tan((x-y)/2)/tan((x+y)/2) gives tan((x-y)/2) = ((k-1)/(k+1)) tan(S/2). (Key point: Tangent relation for x - y) 4. What happens if P lies on the circle through A, B and C? A. The computation becomes more accurate B. Only the coordinate check fails but the answer is unique C. α + β + B = 180°, tan(S/2) is infinite and there is no unique solution D. The sign of the bearing of AP reverses but the solution is unique Answer: C. On the danger circle α + β + B = 180°, so S = 180° and tan(S/2) is infinite; the problem has no unique solution, so P should be chosen away from it. 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