With P outside triangle ABC, ∠APB = 45° and ∠BPC = 52°, what is ∠CPA and why?

7°, as ray PA lies between PC and PB so 52° - 45°

83°, as 360° - 45° - 52° - 180°

When the angles are taken as directed (clockwise positive), what do the three angles at P come to?

+45°, -52° and +7°, summing to 0°

+45°, +52° and +7°, summing to 104°

+45°, +52° and +263°, summing to 360°

-45°, -52° and -7°, summing to -104°

Here K2 and K3 and ΣK are negative. What does that mean for the computation?

The same weighted-mean formula is still used; the signs come from P lying outside the triangle

The method fails and a different method must be used

The signs are dropped and absolute values are used

Only the stations with positive K are used

2080 Chaitra · 8 marks

Tienstra's method, P from A, B, C (APB=45°)

Short answer

Using Tienstra's method with signed observed angles (+45°, -52°, +7°) and interior angles of triangle ABC from coordinates, the computed constants yield P = (18900.471 m E, 15046.510 m N).

The followings are the coordinates of their known stations whose direction are observed from unfixed instrument station P.

Stations	A	B	C
Easting (m)	5000.000	10000.000	15000.000
Northing (m)	1000.000	15000.000	10000.000

If observed horizontal angle APB = 45°00'00" and angle BPC = 52°00'00", determine the coordinates of resection point P by Tienstra method.

The northing of A is taken as printed, 1000.000 m. P lies outside the triangle ABC. The rays from P are in the order C, A, B: $\angle APB = 45^\circ$ , $\angle BPC = 52^\circ$ , and therefore $\angle CPA = 52^\circ - 45^\circ = 7^\circ$ (ray PA lies between PC and PB). Taken as directed angles (clockwise positive), they are $+45^\circ$ , $-52^\circ$ and $+7^\circ$ (sum 0). This is the only position of P that fits the observations.

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: $\alpha = \angle S_2PS_3$ , $\beta = \angle S_3PS_1$ , $\gamma = \angle S_1PS_2$ ( $\alpha+\beta+\gamma = 360^\circ$ , or $0^\circ$ when directed angles of mixed sign are used, as here). 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

Station	Easting (m)	Northing (m)
A	5000.000	1000.000
B	10000.000	15000.000
C	15000.000	10000.000

Angles at P: $\angle APB = \gamma = +45°00'00"$ , $\angle BPC = \alpha = -52°00'00"$ , $\angle CPA = \beta = +7°00'00"$ (sum = 0°00'00").

Step 1: Angles of the triangle ABC from the coordinates

Side	Bearing	Length (m)
AB	19°39'14"	14866.069
BC	135°00'00"	7071.068
CA	228°00'46"	13453.624

Interior angles: $A$ (at A) $= 28°21'32"$ , $B$ (at B) $= 64°39'14"$ , $C$ (at C) $= 86°59'14"$ (sum 180°00'00").

Step 2: Tienstra constants

\begin{aligned} K_1 &= \frac{1}{\cot 28°21'32" - \cot(-52^\circ)} = 0.37966 \\ K_2 &= \frac{1}{\cot 64°39'14" - \cot 7^\circ} = -0.13037 \\ K_3 &= \frac{1}{\cot 86°59'14" - \cot 45^\circ} = -1.05556 \end{aligned}

$\Sigma K = -0.80626$

Step 3: Coordinates of P

E_P = \frac{(0.37966)(5000.000) + (-0.13037)(10000.000) + (-1.05556)(15000.000)}{-0.80626} = 18900.471\ \text{m}

N_P = \frac{(0.37966)(1000.000) + (-0.13037)(15000.000) + (-1.05556)(10000.000)}{-0.80626} = 15046.510\ \text{m}

Check: the clockwise angles computed back from these coordinates are A to B = 45°00'00", B to C = 308°00'00" (that is, 52° anticlockwise) and C to A = 7°00'00", agreeing with the observations.

Answer: coordinates of P = (18900.471 E, 15046.510 N).

✓ Answer checked by Claude Opus 5.5 (high effort)

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Tienstra's method, P from A, B, C (APB=45°)

Data

Step 1: Angles of the triangle ABC from the coordinates

Step 2: Tienstra constants

Step 3: Coordinates of P

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