Plotting a Vacant Lot from Bearings and Distances: 18-Geom-A1 Surveying, May 2016, Q6

Question 6: Plotting a Vacant Lot from Bearings and Distances (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 boundary courses of a closed lot:

Course	Bearing	Distance (m)
A→B	$N20^\circ W$	294.50
B→C	$S69^\circ W$	354.50
C→D	$S20^\circ E$	294.50
D→A	$N69^\circ E$	354.50

Find. A scaled plan sketch of the lot with a north arrow, and a check that the figure closes.

Figure 6 — Vacant lot plotted to shape from the four bearings and distances (north arrow shown). Opposite sides are equal and parallel, so the lot is a parallelogram; it closes exactly on $A$.

Approach. Convert each bearing to a computation azimuth, resolve each course into departure ($E$) and latitude ($N$), plot the vertices by running coordinates, and confirm the departures and latitudes each sum to zero (closure). The $1{:}5{,}000$ scale sets the plotted length ($1$ m on the ground $= 0.2$ mm on paper).

Bearings to azimuths. Reading the quadrant of each bearing: $$N20^\circ W \to 340^\circ,\quad S69^\circ W \to 249^\circ,\quad S20^\circ E \to 160^\circ,\quad N69^\circ E \to 069^\circ$$ Opposite courses differ by exactly $180^\circ$ ($340^\circ/160^\circ$ and $249^\circ/069^\circ$), so the boundary is a parallelogram.
Departures and latitudes. With departure $=L\sin\text{Az}$, latitude $=L\cos\text{Az}$:

Course Departure $E$ (m) Latitude $N$ (m)

A→B $-100.725$ $+276.739$

B→C $-330.954$ $-127.041$

C→D $+100.725$ $-276.739$

D→A $+330.954$ $+127.041$

$\Sigma$ $0.000$ $0.000$

Both column sums are zero, so the lot closes exactly — a geometrically consistent boundary.
Plot the vertices and scale. Running coordinates from $A(0,0)$ give $B(-100.7,+276.7)$, $C(-431.7,+149.7)$, $D(-331.0,-127.0)$ m, then back to $A$. Drawn at $1{:}5{,}000$, the long sides ($354.50$ m) plot as $\boxed{70.9\ \text{mm}}$ and the short sides ($294.50$ m) as $58.9$ mm; orient the sheet with the north arrow up and lay each course off its azimuth with a protractor.

Course	Departure $E$ (m)	Latitude $N$ (m)
A→B	$-100.725$	$+276.739$
B→C	$-330.954$	$-127.041$
C→D	$+100.725$	$-276.739$
D→A	$+330.954$	$+127.041$
$\Sigma$	$0.000$	$0.000$

Quantity	Value
Shape	Parallelogram (opposite sides equal & parallel)
Closure in departure / latitude	$0.000$ m / $0.000$ m (exact)
Plotted long side @ $1{:}5{,}000$	$70.9$ mm ($354.50$ m)
Plotted short side @ $1{:}5{,}000$	$58.9$ mm ($294.50$ m)

Concept & Theory

Topic. Plotting a closed boundary from bearings and distances, with a closure check.

Key relations. Quadrant bearing → azimuth; departure $=L\sin\text{Az}$, latitude $=L\cos\text{Az}$; closure requires $\Sigma\text{dep}=\Sigma\text{lat}=0$; paper length $=$ ground length $\times$ scale.

Why this works. A boundary polygon is closed, so the vector sum of its courses is zero — resolving each into orthogonal departure/latitude components lets that condition be checked arithmetically before ink hits paper. Because $N20^\circ W$ and $S20^\circ E$ (and $N69^\circ E$/$S69^\circ W$) are back-bearings of one another, opposite sides are parallel and equal, giving a parallelogram.

Common pitfall. Mixing the E/W sense of a bearing when forming the azimuth (a westerly bearing gives a negative departure), or plotting distances without applying the map scale. Always plot the north arrow first and reference every azimuth to it.

Source. Wolf & Ghilani, Elementary Surveying §7 (angles, azimuths & bearings) and §18 (mapping / plotting); EGBC standards of practice for cadastral sketches.

Question 6 of 7: Plotting a Vacant Lot from Bearings and Distances

Question 6: Plotting a Vacant Lot from Bearings and Distances (20 marks)