Tienstra's method, coordinates of O

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).