Orientation

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. When the unknown station is observed to three known stations it is called the three-point problem.

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.

         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 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 $AB$ $\pm\, x$ (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.

Example of use

If $\alpha + \beta + B$ is close to $180^\circ$, the point is near the danger circle; the observations are then repeated with another selection of stations.

In the three-point resection problem the coordinates of three stations A, B, C are known and the angles $\alpha = \angle APB$ and $\beta = \angle BPC$ are measured at the unknown station P.

         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 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 $AB$ $\pm\, x$ (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.

In hydrographic survey, each depth measured from the boat is useless unless the position of the boat (sounding point) at the time of sounding is known. The usual methods of fixing the sounding points are given below.

1. By range and a time/distance interval (one range line and distance)

A line (range) is marked on the shore by two poles. The boat keeps on this line and soundings are taken at equal intervals of time or at marked distances, the boat's distance from the shore being found from a tag line or by a rope with marks. It is used for narrow rivers.

2. By two angles from the shore (intersection)

Two theodolites are set up at the ends of a measured base line A and B on the shore. At the signal from the boat, both observers read the angles $\alpha$ and $\beta$ to the boat. The position is plotted by intersection. Accuracy is good, but it needs two observers and good visibility.

        P (boat)
       /  \
      /    \
     A ---- B  (shore base)

3. By one angle and a distance (range and angle)

A theodolite at A sights the boat while the distance AP is measured by a tape/tag line or an EDM. The boat is plotted by polar coordinates (angle and distance).

4. By two sextant angles (resection)

Three known marks on the shore (A, B, C) are used. A sextant at the boat measures the angles $\angle APB$ and $\angle BPC$ at the instant of sounding. The point is plotted by the three-point resection with a station pointer (protractor) or by calculation. No shore observers are needed, which is why it is popular for open water.

5. By a range and a cross angle

A range line on the shore and one angle from a theodolite at a known point on the shore.

6. By GPS (DGPS/RTK) or total station

The receiver on the boat gives the coordinates at each sounding and the position is logged together with the echo sounder depth. It is fast and accurate and is the standard modern method.

Plotting

The base line, shore marks and the sounding positions are plotted at the field scale; the depth, after reduction to the datum for the tide level, is written next to each point and isobaths are interpolated.

The method is chosen by the width of the water, distance from the shore, the accuracy needed and the equipment available. Soundings should be located quickly so that the boat does not move between the time of sounding and the time of fixing.

The three-point problem is solved here by Tienstra's method.

Observed angles at P

Readings: L = 90°00'00", M = 230°58'51", N = 313°17'05", L (closing) = 90°00'30".

• $\angle LPM = 230^\circ58'51'' - 90^\circ00'00'' = 140^\circ58'51''$
• $\angle MPN = 313^\circ17'05'' - 230^\circ58'51'' = 82^\circ18'14''$
• $\angle NPL = 90^\circ00'30'' + 360^\circ - 313^\circ17'05'' = 136^\circ43'25''$

The sum is 360°00'30", so the closing error of 30" is shared equally (−10" to each angle): $\angle LPM = 140^\circ58'41''$, $\angle MPN = 82^\circ18'04''$, $\angle NPL = 136^\circ43'15''$.

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

Angles at P: $\angle LPM = \gamma = 140°58'41"$, $\angle MPN = \alpha = 82°18'04"$, $\angle NPL = \beta = 136°43'15"$ (sum = 360°00'00").

Step 1: Angles of the triangle LMN from the coordinates

Interior angles: $A$ (at L) $= 44°43'08"$, $B$ (at M) $= 65°51'49"$, $C$ (at N) $= 69°25'04"$ (sum 180°00'00").

Step 2: Tienstra constants

$$\begin{aligned} K_1 &= \frac{1}{\cot 44°43' - \cot 82°18'04"} = 1.14328 \\ K_2 &= \frac{1}{\cot 65°51' - \cot 136°43'15"} = 0.66224 \\ K_3 &= \frac{1}{\cot 69°25' - \cot 140°58'41"} = 0.62133 \end{aligned}$$

$\Sigma K = 2.42684$

Step 3: Coordinates of P

$$E_P = \frac{(1.14328)(2880.240) + (0.66224)(3820.600) + (0.62133)(3010.400)}{2.42684} = 3170.170\ \text{m}$$ $$N_P = \frac{(1.14328)(8760.120) + (0.66224)(8000.250) + (0.62133)(7588.800)}{2.42684} = 8252.881\ \text{m}$$

Check: the angles computed back from these coordinates are 140°58'41", 82°18'04" and 136°43'15", equal to the observed angles.

Answer: coordinates of P = (3170.170 E, 8252.881 N).

The observed angles are $\angle NOP = 140^\circ04'45''$, $\angle POG = 99^\circ25'48''$, $\angle GON = 120^\circ28'57''$. Their sum is 359°59'30" = 359°59'30", a misclosure of 30". It is shared equally (+10" to each angle): $\angle NOP = 140^\circ04'55''$, $\angle POG = 99^\circ25'58''$, $\angle GON = 120^\circ29'07''$.

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

Angles at O: $\angle NOP = \gamma = 140°04'55"$, $\angle POG = \alpha = 99°25'58"$, $\angle GON = \beta = 120°29'07"$ (sum = 360°00'00").

Step 1: Angles of the triangle NPG from the coordinates

Interior angles: $A$ (at N) $= 66°28'51"$, $B$ (at P) $= 67°38'08"$, $C$ (at G) $= 45°53'01"$ (sum 180°00'00").

Step 2: Tienstra constants

$$\begin{aligned} K_1 &= \frac{1}{\cot 66°28' - \cot 99°25'58"} = 1.66294 \\ K_2 &= \frac{1}{\cot 67°38' - \cot 120°29'07"} = 0.99986 \\ K_3 &= \frac{1}{\cot 45°53' - \cot 140°04'55"} = 0.46193 \end{aligned}$$

$\Sigma K = 3.12472$

Step 3: Coordinates of O

$$E_O = \frac{(1.66294)(352836.105) + (0.99986)(342615.318) + (0.46193)(329189.716)}{3.12472} = 346069.994\ \text{m}$$ $$N_O = \frac{(1.66294)(3066097.505) + (0.99986)(3050525.416) + (0.46193)(3070164.918)}{3.12472} = 3061715.988\ \text{m}$$

Check: the angles computed back from these coordinates are 140°04'55", 99°25'58" and 120°29'07", equal to the observed angles.

Answer: coordinates of O = (346069.994 E, 3061715.988 N).

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

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

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).

The horizontal angles are the means of face left and face right: at S, $\angle DSO = 75^\circ03'22''$ (both faces agree); at D, $\angle ODS = 68^\circ36'18''$ (both faces agree). O lies south of the line SD. Coordinates are (Easting, Northing).

        S ------------- D
         \            /
          \ 75°      / 68°
           \        /
            \      /
             \    /
              \  /
               O

Step 1: Length and bearing of the base SD

$$\Delta E = 2137.342,\quad \Delta N = -1565.199$$ $$SD = \sqrt{\Delta E^2 + \Delta N^2} = 2649.166\ \text{m}, \qquad \text{bearing of SD} = 126°12'57"\ (\text{south-east})$$

Step 2: Triangle SDO

$\angle SOD = 180^\circ - 75^\circ03'22'' - 68^\circ36'18'' = 36°20'20"$

By the sine rule:

$$SO = \frac{SD\sin 68^\circ36'18''}{\sin 36°20'20"} = 4162.625\ \text{m}, \qquad DO = \frac{SD\sin 75^\circ03'22''}{\sin 36°20'20"} = 4319.508\ \text{m}$$

Step 3: Bearings and coordinates

O is south of SD, i.e. on the right-hand side when facing from S to D, so the bearing of SO = bearing of SD + 75°03'22" $= 201°16'19"$.

$$E_O = E_S + SO\sin\theta = 627464.712 + 4162.625\sin 201°16'19" = 625954.540\ \text{m}$$ $$N_O = N_S + SO\cos\theta = 3066928.474 + 4162.625\cos 201°16'19" = 3063049.450\ \text{m}$$

Check from D: bearing of DO $= 306°12'57" - 68^\circ36'18'' = 237°36'39"$, giving $E = 625954.540$ m and $N = 3063049.450$ m (the same).

Answer: coordinates of O = (625954.540 E, 3063049.450 N).

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

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

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).

The three-point problem is solved by Tienstra's method.

Observed angles at O

Readings: A = 90°00'00", B = 230°58'51", C = 313°17'05", A (closing) = 90°00'30".

• $\angle AOB = 140^\circ58'51''$, $\angle BOC = 82^\circ18'14''$, $\angle COA = 136^\circ43'25''$ (sum 360°00'30")
• The closing error of 30" is shared equally (−10" each): $\angle AOB = 140^\circ58'41''$, $\angle BOC = 82^\circ18'04''$, $\angle COA = 136^\circ43'15''$.

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

Angles at O: $\angle AOB = \gamma = 140°58'41"$, $\angle BOC = \alpha = 82°18'04"$, $\angle COA = \beta = 136°43'15"$ (sum = 360°00'00").

Step 1: Angles of the triangle ABC from the coordinates

Interior angles: $A$ (at A) $= 47°14'34"$, $B$ (at B) $= 69°55'56"$, $C$ (at C) $= 62°49'30"$ (sum 180°00'00").

Step 2: Tienstra constants

$$\begin{aligned} K_1 &= \frac{1}{\cot 47°14' - \cot 82°18'04"} = 1.26672 \\ K_2 &= \frac{1}{\cot 69°55' - \cot 136°43'15"} = 0.70065 \\ K_3 &= \frac{1}{\cot 62°49' - \cot 140°58'41"} = 0.57231 \end{aligned}$$

$\Sigma K = 2.53967$

Step 3: Coordinates of O

$$E_O = \frac{(1.26672)(2876.240) + (0.70065)(3810.800) + (0.57231)(2959.390)}{2.53967} = 3152.804\ \text{m}$$ $$N_O = \frac{(1.26672)(8754.110) + (0.70065)(7997.250) + (0.57231)(7487.090)}{2.53967} = 8259.788\ \text{m}$$

Check: the angles computed back from these coordinates are 140°58'41", 82°18'04" and 136°43'15", equal to the observed angles.

Answer: coordinates of O = (3152.804 E, 8259.788 N).

Difference between intersection and resection

Intersection is the method of finding the coordinates of an unknown point P by measuring the horizontal angles to P from two (or more) stations of known coordinates. The instrument is set up at the known stations, not at P.

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.

Three-point resection method

The coordinates of three known stations A, B, C are given; the angles $\alpha = \angle APB$ and $\beta = \angle BPC$ are measured at the unknown point P.

1. Compute the sides $c = AB$, $a = BC$ and the angle $B$ at the middle station from the coordinates.
2. $S = x + y = 360^\circ - (\alpha + \beta + B)$, where $x = \angle PAB$ and $y = \angle PCB$.
3. $k = \dfrac{a\sin\alpha}{c\sin\beta}$ and $\tan\dfrac{x-y}{2} = \dfrac{k-1}{k+1}\tan\dfrac{S}{2}$; hence $x$ and $y$.
4. $AP = \dfrac{c\sin(180^\circ - \alpha - x)}{\sin\alpha}$, the bearing of AP = bearing of AB $\pm x$, and $E_P = E_A + AP\sin\theta,\; N_P = N_A + AP\cos\theta$.
5. Check by computing P from C.

A graphical (tracing paper) solution also exists, and Tienstra's formula gives the coordinates directly as the weighted mean of the three stations.

The northing of Harisiddhi is printed as 6065364.275; it is taken as 3065364.275 (a misprint, since the other stations lie near 3 06x xxx N). O lies outside the triangle SDC. $\angle SOD = 40^\circ20'21''$, $\angle DOC = 98^\circ56'41''$, $\angle COS = 220^\circ42'58''$ (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

Angles at O: $\angle SOD = \gamma = 40°20'21"$, $\angle DOC = \alpha = 98°56'41"$, $\angle COS = \beta = 220°42'58"$ (sum = 360°00'00").

Step 1: Angles of the triangle SDC from the coordinates

Interior angles: $A$ (at S) $= 52°14'20"$, $B$ (at D) $= 99°05'18"$, $C$ (at C) $= 28°40'22"$ (sum 180°00'00").

Step 2: Tienstra constants

$$\begin{aligned} K_1 &= \frac{1}{\cot 52°14' - \cot 98°56'41"} = 1.07297 \\ K_2 &= \frac{1}{\cot 99°05' - \cot 220°42'58"} = -0.75648 \\ K_3 &= \frac{1}{\cot 28°40' - \cot 40°20'21"} = 1.53593 \end{aligned}$$

$\Sigma K = 1.85242$

Step 3: Coordinates of O

$$E_O = \frac{(1.07297)(627465.718) + (-0.75648)(629603.054) + (1.53593)(627612.753)}{1.85242} = 626714.801\ \text{m}$$ $$N_O = \frac{(1.07297)(3066929.474) + (-0.75648)(3065364.275) + (1.53593)(3061479.468)}{1.85242} = 3063049.803\ \text{m}$$

Check: the angles computed back from these coordinates are 40°20'21", 98°56'41" and 220°42'58", equal to the observed angles.

Answer: coordinates of O = (626714.801 E, 3063049.803 N).

Definitions

Intersection is the method of finding the coordinates of an unknown point P by measuring the horizontal angles to P from two (or more) stations of known coordinates. The instrument is set up at the known stations, not at P.

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.

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.

         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 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 $AB$ $\pm\, x$ (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.

Resection

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.

Two-point problem

The two-point problem is used when only two stations of known coordinates (A and B) are available. Two points alone cannot fix one unknown station by angles at that station, so two unknown stations (C and D, which are intervisible) are set up. At each of them the angles to A and B and to the other unknown station are observed. The angles of the quadrilateral give the angles at A and B (Hansen's method), and the sine rule then gives the distances and the coordinates of both stations.

Intersection

Intersection is the method of finding the coordinates of an unknown point P by measuring the horizontal angles to P from two (or more) stations of known coordinates. The instrument is set up at the known stations, not at P.

Numerical (coordinates of P from the angles at A and B)

The observed angles are at the known stations A and B, so P is found by intersection. Coordinates are (Easting, Northing): A = (3932 E, 7492 N), B = (2960 E, 7487 N). The angles are $\angle BAP = 44^\circ52'36''$ at A and $\angle ABP = 75^\circ33'22''$ at B. P is taken on the north side of AB (the south side would give the mirror image).

$$AB = \sqrt{(2960-3932)^2 + (7487-7492)^2} = 972.013\ \text{m}, \quad \text{bearing of AB} = 269°42'19"$$

$\angle APB = 180^\circ - 44^\circ52'36'' - 75^\circ33'22'' = 59°34'02"$

$$AP = \frac{AB\sin 75^\circ33'22''}{\sin 59°34'02"} = 1091.700\ \text{m},\qquad BP = \frac{AB\sin 44^\circ52'36''}{\sin 59°34'02"} = 795.426\ \text{m}$$

Bearing of AP = bearing of AB + 44°52'36" $= 314°34'55"$ (AB runs almost due west, so turning clockwise puts P to the north).

$$E_P = 3932 + 1091.700\sin 314°34'55" = 3154.440\ \text{m},\qquad N_P = 7492 + 1091.700\cos 314°34'55" = 8258.295\ \text{m}$$

Check from B: bearing of BP $= 89°42'19" - 75^\circ33'22'' = 14°08'57"$ gives $E = 3154.440$, $N = 8258.295$ (same).

Answer: P = (3154.440 E, 8258.295 N).

Note on the data. With A = (5000 E, 10000 N), B = (10000, 15000) and C = (15000, 10000), triangle ABC is right-angled and isosceles with $\angle ACB = 45^\circ$. Then $\angle APB = 45^\circ$ is possible only for P on the circle through A, B and C (the danger circle), or at C itself, and for such points $\angle BPC$ cannot be $52^\circ$; so no true solution exists. The northing of A is therefore taken as 1000.000 m (as in the companion problem), which gives a proper resection. P lies outside the triangle.

Angles at P: $\angle APB = 45^\circ$, $\angle BPC = 52^\circ$; the rays are in the order C, A, B, so $\angle CPA = 52^\circ - 45^\circ = 7^\circ$.

Triangle ABC from coordinates (E, N)

Interior angles: $A = 28°21'32"$, $B = 64°39'14"$, $C = 86°59'14"$.

Tienstra's method (directed angles: $\gamma = \angle APB = +45^\circ$, $\alpha = \angle BPC = -52^\circ$, $\beta = \angle CPA = +7^\circ$)

$$K_A = \frac{1}{\cot A - \cot\alpha} = 0.37966,\quad K_B = \frac{1}{\cot B - \cot\beta} = -0.13037,\quad K_C = \frac{1}{\cot C - \cot\gamma} = -1.05556$$ $$E_P = \frac{\sum K E}{\sum K} = 18900.471\ \text{m},\qquad N_P = \frac{\sum K N}{\sum K} = 15046.510\ \text{m}$$

(i) Lengths and bearings

$$\text{Length} = \sqrt{\Delta E^2 + \Delta N^2}, \qquad \text{bearing} = \tan^{-1}\frac{\Delta E}{\Delta N}\ (\text{quadrant corrected})$$

Check: bearing PB − bearing PA $= 45°00'00"$ (= 45°) and bearing PB − bearing PC $= 52°00'00"$ (= 52°). The bearings of AP, BP, CP (from the stations to P) are these bearings $\pm 180^\circ$.

(ii) Coordinates of P

Answer: P = (18900.471 E, 15046.510 N). AP = 19761.769 m, BP = 8900.592 m, CP = 6378.161 m; bearings from P: to A 224°42'02", to B 269°42'02", to C 217°42'02".