Coordinates of O from Swoyambhu, Harisiddhi, Chovar

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).