Coordinates and bearings of P from A, B, C

Note on the data. With A = (5000 E, 10000 N), B = (10000, 15000) and C = (15000, 10000), triangle ABC is right-angled at B and isosceles ($\angle A = \angle C = 45^\circ$). The points that see AB at 45° on C's side of AB lie on the circle through A, B and C, and from every such point BC is seen at 45° or 135°, never 52°. Checking all other positions (including the far side of AB) also gives no point where $\angle APB = 45^\circ$ and $\angle BPC = 52^\circ$ at the same time. So the data as printed has no solution. The northing of A is therefore taken as 1000.000 m (as in the companion problem), which gives a proper resection. P then lies outside the triangle.

Angles at P: $\angle APB = 45^\circ$, $\angle BPC = 52^\circ$; the rays are in the order C, A, B, so $\angle CPA = 52^\circ - 45^\circ = 7^\circ$.

Triangle ABC from coordinates (E, N)

Interior angles: $A = 28°21'32"$, $B = 64°39'14"$, $C = 86°59'14"$.

Tienstra's method (directed angles: $\gamma = \angle APB = +45^\circ$, $\alpha = \angle BPC = -52^\circ$, $\beta = \angle CPA = +7^\circ$)

$$K_A = \frac{1}{\cot A - \cot\alpha} = 0.37966,\quad K_B = \frac{1}{\cot B - \cot\beta} = -0.13037,\quad K_C = \frac{1}{\cot C - \cot\gamma} = -1.05556$$ $$E_P = \frac{\sum K E}{\sum K} = 18900.471\ \text{m},\qquad N_P = \frac{\sum K N}{\sum K} = 15046.510\ \text{m}$$

(i) Lengths and bearings

$$\text{Length} = \sqrt{\Delta E^2 + \Delta N^2}, \qquad \text{bearing} = \tan^{-1}\frac{\Delta E}{\Delta N}\ (\text{quadrant corrected})$$

Check: bearing PB − bearing PA $= 45°00'00"$ and bearing PB − bearing PC $= 52°00'00"$, equal to the observed angles. The bearings of AP, BP and CP (from the stations to P) are the bearings from P $\pm 180^\circ$.

(ii) Coordinates of P

Answer (with $N_A$ = 1000 m): P = (18900.471 E, 15046.510 N). AP = 19761.769 m (bearing 44°42'02"), BP = 8900.592 m (bearing 89°42'02"), CP = 6378.161 m (bearing 37°42'02").