Resection point O inside triangle N, P, G

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", 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 the unknown point 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

$$\begin{aligned} K_1 &= \frac{1}{\cot 66°28'51" - \cot 99°25'58"} = 1.66294 \\ K_2 &= \frac{1}{\cot 67°38'08" - \cot 120°29'07"} = 0.99986 \\ K_3 &= \frac{1}{\cot 45°53'01" - \cot 140°04'55"} = 0.46193 \end{aligned}$$

$\Sigma K = 3.12472$

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 adjusted observed angles.

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