Tienstra's method, O from P, Q, R

Coordinates are given as Northing, Easting; they are used here as (E, N): P = (24078.31 E, 29236.48 N), Q = (26266.48 E, 31493.20 N), R = (28377.67 E, 29661.04 N).

Observed: $\angle POQ = 142^\circ43'32''$, $\angle QOR = \angle ROQ = 92^\circ19'52''$, $\angle ROP = \angle POR = 124^\circ56'26''$. The sum is 359°59'50" (misclosure 10"), shared equally (+3.3" to each): 142°43'35", 92°19'55", 124°56'29".

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: $\angle POQ = \gamma = 142°43'35"$, $\angle QOR = \alpha = 92°19'55"$, $\angle ROP = \beta = 124°56'29"$ (sum = 360°00'00").

Step 1: Angles of the triangle PQR from the coordinates

Interior angles: $A$ (at P) $= 40°14'38"$, $B$ (at Q) $= 93°09'50"$, $C$ (at R) $= 46°35'32"$ (sum 180°00'00").

Step 2: Tienstra constants

$$\begin{aligned} K_1 &= \frac{1}{\cot 40°14' - \cot 92°19'55"} = 0.81818 \\ K_2 &= \frac{1}{\cot 93°09' - \cot 124°56'29"} = 1.55422 \\ K_3 &= \frac{1}{\cot 46°35' - \cot 142°43'35"} = 0.44251 \end{aligned}$$

$\Sigma K = 2.81490$

Step 3: Coordinates of O

$$E_O = \frac{(0.81818)(24078.310) + (1.55422)(26266.480) + (0.44251)(28377.670)}{2.81490} = 25962.348\ \text{m}$$ $$N_O = \frac{(0.81818)(29236.480) + (1.55422)(31493.200) + (0.44251)(29661.040)}{2.81490} = 30549.246\ \text{m}$$

Check: the angles computed back from these coordinates are 142°43'35", 92°19'55" and 124°56'29", equal to the observed angles.

Answer: coordinates of O = (25962.348 E, 30549.246 N).