Why is 360° added when finding the angle NPL from the readings 313°17'05" (N) and 90°00'30" (closing L)?

Because the closing reading is smaller than the previous reading, so the angle is measured clockwise through zero

Because the closing error must be removed

Because the face-left and face-right readings are combined

Because the angle must always exceed 180°

The three angles at P sum to 360°00'30". What is done before applying Tienstra's formula?

The 30" is shared equally, subtracting 10" from each angle

The 30" is added to the largest angle only

The 30" is ignored since it is small

The readings are repeated with a different formula

In Tienstra's formula the constant for station L uses interior angle A (at L) together with which observed angle at P?

The angle MPN, subtended by the side opposite L

The sum of all three angles at P

How are the coordinates of P obtained once K1, K2 and K3 are known?

As the weighted mean E_P = (K1E1 + K2E2 + K3E3)/(K1+K2+K3), and likewise for N

As the plain average of the three stations' coordinates

By adding the K values to the station coordinates

By taking the coordinates of the station with the largest K

2079 Jestha · 6 marks

Resection point P from L, M, N

Short answer

Using Tienstra's method with observed angles at P adjusted to ∠LPM=140°58'41", ∠MPN=82°18'04", ∠NPL=136°43'15", and triangle angles computed from coordinates, the resection point coordinates are P (3170.170 mE, 8252.881 mN).

Station L, M and N have the following respective coordinates (2880.24 mE, 8760.12 mN), (3820.60 mE, 8000.25 mN) and (3010.40 mE, 7588.80 mN) respectively. Station P is resection point and following observations were recorded. Determine the coordinate of P.

Stn	Sighted to	HCR
P	L	90°00'00"
	M	230°58'51"
	N	313°17'05"
	L	90°00'30"

The three-point problem is solved here by Tienstra's method.

Observed angles at P

Readings: L = 90°00'00", M = 230°58'51", N = 313°17'05", L (closing) = 90°00'30".

$\angle LPM = 230^\circ58'51'' - 90^\circ00'00'' = 140^\circ58'51''$
$\angle MPN = 313^\circ17'05'' - 230^\circ58'51'' = 82^\circ18'14''$
$\angle NPL = 90^\circ00'30'' + 360^\circ - 313^\circ17'05'' = 136^\circ43'25''$

The sum is 360°00'30", so the closing error of 30" is shared equally (−10" to each angle): $\angle LPM = 140^\circ58'41''$ , $\angle MPN = 82^\circ18'04''$ , $\angle NPL = 136^\circ43'15''$ .

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

Station	Easting (m)	Northing (m)
L	2880.240	8760.120
M	3820.600	8000.250
N	3010.400	7588.800

Angles at P: $\angle LPM = \gamma = 140°58'41"$ , $\angle MPN = \alpha = 82°18'04"$ , $\angle NPL = \beta = 136°43'15"$ (sum = 360°00'00").

Step 1: Angles of the triangle LMN from the coordinates

Side	Bearing	Length (m)
LM	128°56'25"	1208.999
MN	243°04'37"	908.689
NL	353°39'33"	1178.530

Interior angles: $A$ (at L) $= 44°43'08"$ , $B$ (at M) $= 65°51'49"$ , $C$ (at N) $= 69°25'04"$ (sum 180°00'00").

Step 2: Tienstra constants

\begin{aligned} K_1 &= \frac{1}{\cot 44°43' - \cot 82°18'04"} = 1.14328 \\ K_2 &= \frac{1}{\cot 65°51' - \cot 136°43'15"} = 0.66224 \\ K_3 &= \frac{1}{\cot 69°25' - \cot 140°58'41"} = 0.62133 \end{aligned}

$\Sigma K = 2.42684$

Step 3: Coordinates of P

E_P = \frac{(1.14328)(2880.240) + (0.66224)(3820.600) + (0.62133)(3010.400)}{2.42684} = 3170.170\ \text{m}

N_P = \frac{(1.14328)(8760.120) + (0.66224)(8000.250) + (0.62133)(7588.800)}{2.42684} = 8252.881\ \text{m}

Check: the angles computed back from these coordinates are 140°58'41", 82°18'04" and 136°43'15", equal to the observed angles.

Answer: coordinates of P = (3170.170 E, 8252.881 N).

✓ Answer checked by Claude Opus 5.5 (high effort)

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