Location of Station C by Intersection: 18-Geom-A1 Surveying, May 2019, Q6

Question 6: Location of Station C by Intersection (10 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. $X_A = 433\,191.050$ m, $Y_A = 158\,893.500$ m; line A→B has azimuth $\theta_{AB} = 235^\circ20'32''$ and length $d_{AB} = 895.425$ m; the measured angles are $\angle BAC = 80^\circ27'35.8''$ and $\angle CBA = 54^\circ14'37.8''$. $X$ is easting and $Y$ is northing.

Find. The coordinates $(X_C, Y_C)$ of station C.

Intersection triangle A–B–C plotted to scale from the computed coordinates (north up); the dashed triangle is the mirror solution on the other side of AB.

Assumption — side of AB. No sketch is printed, so the side of AB on which C lies must be assumed. The standard surveying reading of an angle named ∠BAC is the angle at A turned clockwise (to the right) from B to C, and ∠CBA is turned clockwise at B from C to A. This gives $\theta_{AC} = \theta_{AB} + \angle BAC$ and $\theta_{BC} = \theta_{BA} - \angle CBA$, which is the boxed solution. If the angles were turned counter-clockwise, C falls on the other side of AB at $(433625.022,\ 157967.815)$; the working is identical apart from the signs.

Approach. Fix B from A using the given azimuth and distance, find the third angle of the triangle, get the sides AC and BC by the sine rule, then compute C from A and check it independently from B.

Coordinates of B. With $\Delta X = d\sin\theta$ and $\Delta Y = d\cos\theta$: $$ X_B = 433191.050 + 895.425\sin(235^\circ20'32'') = 432454.506\text{ m} $$ $$ Y_B = 158893.500 + 895.425\cos(235^\circ20'32'') = 158384.296\text{ m} $$
Third angle. $$ \angle ACB = 180^\circ - 80^\circ27'35.8'' - 54^\circ14'37.8'' = 45^\circ17'46.4''. $$
Sides by the sine rule. $$ AC = d_{AB}\frac{\sin\angle CBA}{\sin\angle ACB} = 895.425\,\frac{\sin 54^\circ14'37.8''}{\sin 45^\circ17'46.4''} = 1022.363\text{ m}, \qquad BC = d_{AB}\frac{\sin\angle BAC}{\sin\angle ACB} = 1242.402\text{ m}. $$
Azimuth A→C and coordinates of C. $\theta_{AC} = 235^\circ20'32'' + 80^\circ27'35.8'' = 315^\circ48'07.8''$: $$ X_C = 433191.050 + 1022.363\sin(315^\circ48'07.8'') = 433191.050 -712.728 \;\Rightarrow\; \boxed{X_C = 432478.322\text{ m}} $$ $$ Y_C = 158893.500 + 1022.363\cos(315^\circ48'07.8'') = 158893.500 +732.969 \;\Rightarrow\; \boxed{Y_C = 159626.469\text{ m}} $$
Independent check from B. $\theta_{BA} = 55^\circ20'32''$, so $\theta_{BC} = \theta_{BA} - \angle CBA = 1^\circ05'54.2''$. Then $$ X_C = 432454.506 + 1242.402\sin\theta_{BC} = 432478.322, \qquad Y_C = 158384.296 + 1242.402\cos\theta_{BC} = 159626.469. $$ The two rays meet at the same point, which confirms the intersection.

Quantity	Value
Coordinates of B	$(432454.506,\ 158384.296)$ m
Angle at C, $\angle ACB$	$45^\circ17'46.4''$
Sides AC / BC	$1022.363$ / $1242.402$ m
Azimuths A→C / B→C	$315^\circ48'07.8''$ / $1^\circ05'54.2''$
Coordinates of C (angles to the right)	$(432478.322,\ 159626.469)$ m
Mirror solution (angles turned left)	$(433625.022,\ 157967.815)$ m

Concept & Theory

Topic. Fixing a new station by angular intersection from two known control points.

Key relations. Sine rule $AC/\sin B = BC/\sin A = AB/\sin C$; coordinate propagation $\Delta X = d\sin\theta$, $\Delta Y = d\cos\theta$; azimuth carry $\theta_{AC} = \theta_{AB} \pm \angle A$.

Why this works. A known baseline and the two angles at its ends fix a unique triangle. The sine rule gives the unknown sides, and carrying the baseline azimuth through each measured angle gives the directions to C. Computing C from both A and B is a built-in closure check.

Common pitfall. Putting C on the wrong side of the baseline, which gives the mirror point, and swapping $\sin$ and $\cos$ for easting and northing. Check the answer by computing from the second point.

Source. Wolf & Ghilani, Elementary Surveying, 15th ed., Ch. 11 (coordinate geometry: intersection) and Ch. 10 (coordinate computations).

Question 6 of 9: Location of Station C by Intersection

Question 6: Location of Station C by Intersection (10 marks)