How is the problem split into parts?

O is found from known stations P, W, D by Tienstra, then A is located from O by bearing and distance

A is found first, then O from A

Both are found at once by Tienstra

O is assumed and A is found by sine rule only

Which observed angles at O are used with Tienstra's formula for P, W, D?

∠POW, ∠WOD and ∠DOP, which total 360°

How is the bearing of OA found?

Bearing of OW plus 42°18'14" (clockwise) = 153°31'50"

Which expressions give the coordinates of A?

E_A = E_O + OA sinθ, N_A = N_O + OA cosθ

E_A = E_O + OA cosθ, N_A = N_O + OA sinθ

E_A = E_O - OA sinθ, N_A = N_O - OA cosθ

2079 Chaitra · 8 marks

Coordinates of instrument station O and station A

Short answer

O is determined by Tienstra's method as (3170.205 m E, 8252.871 m N). From bearing OW and angle WOA, bearing OA is computed; using length OA=140.23 m, A = (3232.708 m E, 8127.341 m N).

The following are the three known points. The information regarding the three points and one traverse leg are given below. Compute the coordinates of instrument station and the traverse station A.

Instrument station	Sighted to	CW observed angles	Northing (m)	Easting (m)	Remarks
O	P	∠POW = 140°58'51"	8760.12	2880.24
	W	∠WOD = 82°18'14"	8000.25	3820.60
	A	∠WOA = 42°18'14"	?	?	Length (OA) = 140.23 m
	D	∠DOP = 136°42'55"	7588.80	3010.40
	P		8760.12	2880.24

The coordinates are written as (Easting, Northing): P = (2880.24 E, 8760.12 N), W = (3820.60 E, 8000.25 N), D = (3010.40 E, 7588.80 N). Observed clockwise angles at O: $\angle POW = 140^\circ58'51''$ , $\angle WOD = 82^\circ18'14''$ , $\angle DOP = 136^\circ42'55''$ (sum 360°).

Part 1: Coordinates of the instrument station O (Tienstra's method)

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	2880.240	8760.120
W	3820.600	8000.250
D	3010.400	7588.800

Angles at O: $\angle POW = \gamma = 140°58'51"$ , $\angle WOD = \alpha = 82°18'14"$ , $\angle DOP = \beta = 136°42'55"$ (sum = 360°00'00").

Step 1: Angles of the triangle PWD from the coordinates

Side	Bearing	Length (m)
PW	128°56'25"	1208.999
WD	243°04'37"	908.689
DP	353°39'33"	1178.530

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

Step 2: Tienstra constants

\begin{aligned} K_1 &= \frac{1}{\cot 44°43' - \cot 82°18'14"} = 1.14321 \\ K_2 &= \frac{1}{\cot 65°51' - \cot 136°42'55"} = 0.66233 \\ K_3 &= \frac{1}{\cot 69°25' - \cot 140°58'51"} = 0.62128 \end{aligned}

$\Sigma K = 2.42682$

Step 3: Coordinates of O

E_O = \frac{(1.14321)(2880.240) + (0.66233)(3820.600) + (0.62128)(3010.400)}{2.42682} = 3170.205\ \text{m}

N_O = \frac{(1.14321)(8760.120) + (0.66233)(8000.250) + (0.62128)(7588.800)}{2.42682} = 8252.871\ \text{m}

Check: the angles computed back from these coordinates are 140°58'51", 82°18'14" and 136°42'55", equal to the observed angles.

Coordinates of O = (3170.205 E, 8252.871 N).

Part 2: Coordinates of the traverse station A

$\angle WOA = 42^\circ18'14''$ is measured clockwise from OW (A lies between W and D, since $\angle WOD = 82^\circ18'14''$ ), and $OA = 140.23$ m.

Bearing of OW (from O to W) $= 111°13'36"$

Bearing of OA = bearing of OW + 42°18'14" $= 153°31'50"$

E_A = E_O + OA\sin\theta = 3170.205 + 140.23\sin 153°31'50" = 3232.708\ \text{m}

N_A = N_O + OA\cos\theta = 8252.871 + 140.23\cos 153°31'50" = 8127.341\ \text{m}

Answer: O = (3170.205 E, 8252.871 N); A = (3232.708 E, 8127.341 N).

✓ Answer checked by Claude Opus 5.5 (high effort)

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