Tienstra's method, X from Swayambhu, Dharara, Chovar

X lies outside the triangle SDC: $\angle SXD = 41^\circ20'21''$, $\angle DXC = 97^\circ56'41''$, $\angle CXS = 220^\circ42'58''$ (sum 360°). The angle above 180° is used as measured clockwise.

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 X: $\angle SXD = \gamma = 41°20'21"$, $\angle DXC = \alpha = 97°56'41"$, $\angle CXS = \beta = 220°42'58"$ (sum = 360°00'00").

Step 1: Angles of the triangle SDC from the coordinates

Interior angles: $A$ (at S) $= 52°14'19"$, $B$ (at D) $= 99°04'57"$, $C$ (at C) $= 28°40'45"$ (sum 180°00'00").

Step 2: Tienstra constants

$$\begin{aligned} K_1 &= \frac{1}{\cot 52°14' - \cot 97°56'41"} = 1.09390 \\ K_2 &= \frac{1}{\cot 99°04' - \cot 220°42'58"} = -0.75654 \\ K_3 &= \frac{1}{\cot 28°40' - \cot 41°20'21"} = 1.44631 \end{aligned}$$

$\Sigma K = 1.78367$

Step 3: Coordinates of X

$$E_X = \frac{(1.09390)(627464.718) + (-0.75654)(629602.054) + (1.44631)(627611.753)}{1.78367} = 626677.395\ \text{m}$$ $$N_X = \frac{(1.09390)(3066928.474) + (-0.75654)(3065363.275) + (1.44631)(3061479.468)}{1.78367} = 3063173.956\ \text{m}$$

Check: the angles computed back from these coordinates are 41°20'21", 97°56'41" and 220°42'58", equal to the observed angles.

Answer: coordinates of X = (626677.395 E, 3063173.956 N).