Weighted Adjustment of a Triangle's Angles: 18-Geom-A1 Surveying, December 2013, Q4

Question 4: Weighted Adjustment of a Triangle's Angles (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four repeated observations of each angle of a plane triangle:

Set	Angle A	Angle B	Angle C
1	$38^\circ47'54''$	$71^\circ22'33''$	$69^\circ50'08''$
2	$38^\circ47'49''$	$71^\circ22'27''$	$69^\circ50'11''$
3	$38^\circ48'17''$	$71^\circ22'17''$	$69^\circ50'39''$
4	$38^\circ48'13''$	$71^\circ22'17''$	$69^\circ50'15''$

Find. Relative weights of $A,B,C$ and the adjusted angles (to the nearest second) satisfying $A+B+C=180^\circ$.

Figure 4 — Plane triangle (not to scale); the three interior angles must satisfy $A+B+C=180^\circ$.

Approach. Mean each angle, form the closure misclosure against $180^\circ$, derive relative weights from the internal consistency (sample variance) of each set, then distribute the misclosure inversely to weight.

Mean of each angle. $$\bar A = 38^\circ48'03.25'',\quad \bar B = 71^\circ22'23.50'',\quad \bar C = 69^\circ50'18.25''$$ Their sum is $180^\circ00'45.00''$, so the triangle misclosure is $w=+45''$ (plane condition $A+B+C=180^\circ$).
Relative weights from consistency. Each angle is observed the same number of times (4), so weight is governed by the spread. Using the sample variance $s^2$ (arc-sec²) of each set: $$s_A^2 = 191,\quad s_B^2 = 62,\quad s_C^2 = 200$$ $$p_A:p_B:p_C = \tfrac1{191}:\tfrac1{62}:\tfrac1{200} = \boxed{0.199:0.610:0.191}$$ Angle $B$, the most internally consistent set, is the most heavily weighted.
Distribute the misclosure inversely to weight. The correction is proportional to $s^2$, $v_i=-w\,s_i^2/\Sigma s^2$ with $\Sigma s^2=453$: $$v_A=-18.97'',\quad v_B=-6.19'',\quad v_C=-19.83''\qquad(\Sigma=-45'')$$ The unrounded adjusted values are $A=38^\circ47'44.28''$, $B=71^\circ22'17.31''$ and $C=69^\circ49'58.42''$. Rounding each one independently would give a sum of $179^\circ59'59''$, so the one-second rounding deficit goes to the value with the largest fractional part ($C$, $0.42''$), which keeps the rounded set exactly closed. Adjusted angles (to the nearest second): $$\bar A = \boxed{38^\circ47'44''},\quad \bar B = \boxed{71^\circ22'17''},\quad \bar C = \boxed{69^\circ49'59''}$$ Adjusted sum $=180^\circ00'00''$. The best-observed angle ($B$) changes least; the noisier angles absorb most of the correction.

Angle	Weight ratio	Correction	Adjusted value
A	0.199	$-18.97''$	$38^\circ47'44''$
B	0.610	$-6.19''$	$71^\circ22'17''$
C	0.191	$-19.83''$	$69^\circ49'59''$

Concept & Theory

Topic. Weighted adjustment of angle observations under a single geometric condition (angle closure of a plane triangle).

Key relations. Weight $p_i\propto1/s_i^2$; single-condition least-squares correction $v_i=-w\,s_i^2/\Sigma s^2$ (proportional to $1/p_i$); closure $\sum\bar\angle=180^\circ$.

Why this works. Minimising $\sum p_i v_i^2$ subject to $\sum\bar\angle=180^\circ$ gives corrections proportional to $1/p_i$, i.e. to $s_i^2$. Deriving the weights from the observations themselves — as the question demands with "determine relative weights" — pushes the closing error onto the least-trustworthy angles, which is exactly the intended behaviour.

Common pitfall. Splitting the $45''$ equally (that assumes equal precision), or inverting the rule and correcting the low-variance angle the most. Watch the sample-variance divisor $n-1=3$, not $n$.

Source. Ghilani, Adjustment Computations §10–11 (weights & the condition equation); Wolf & Ghilani, Elementary Surveying §16 (introduction to adjustments).

Question 4 of 7: Weighted Adjustment of a Triangle's Angles

Question 4: Weighted Adjustment of a Triangle's Angles (20 marks)