Deflection Angles and Interior Angles from Bearings: 18-Geom-A1 Surveying, May 2016, Q7

Question 7: Deflection Angles and Interior Angles from Bearings (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. An open route traverse $A$–$B$–$C$–$D$–$E$ with line bearings:

Side	Bearing	Azimuth
AB	$N61^\circ24'10''E$	$61^\circ24'10''$
BC	$N88^\circ36'40''E$	$88^\circ36'40''$
CD	$S9^\circ33'32''E$	$170^\circ26'28''$
DE	$N71^\circ10'28''W$	$288^\circ49'32''$

Find. (1) the deflection angle at each intermediate station B, C and D; (2) the interior angles at B and D.

Figure 7 — Open route traverse A→B→C→D→E plotted from the four bearings (north arrow shown); the route deflects to the right at B, C and D.

Approach. Convert each bearing to an azimuth, take the deflection angle at a station as the change in azimuth (forward $-$ back), and obtain each interior angle as $180^\circ$ minus the deflection (equivalently, the angle from the back line reversed to the forward line).

Bearings to azimuths. Applying the quadrant rules ($NE$: Az $=$ bearing; $SE$: Az $=180^\circ-$ bearing; $NW$: Az $=360^\circ-$ bearing): $$\text{Az}_{AB}=61^\circ24'10'',\ \text{Az}_{BC}=88^\circ36'40'',\ \text{Az}_{CD}=170^\circ26'28'',\ \text{Az}_{DE}=288^\circ49'32''$$
Deflection angles (forward azimuth $-$ back azimuth). A positive result is a right (clockwise) deflection: $$\delta_B = \text{Az}_{BC}-\text{Az}_{AB} = 88^\circ36'40''-61^\circ24'10'' = \boxed{27^\circ12'30''\ \text{R}}$$ $$\delta_C = \text{Az}_{CD}-\text{Az}_{BC} = 170^\circ26'28''-88^\circ36'40'' = \boxed{81^\circ49'48''\ \text{R}}$$ $$\delta_D = \text{Az}_{DE}-\text{Az}_{CD} = 288^\circ49'32''-170^\circ26'28'' = \boxed{118^\circ23'04''\ \text{R}}$$
Interior angles at B and D. The interior angle is the angle measured at the station from the reversed back line to the forward line, equal to $180^\circ-\delta$ for a right deflection: $$\angle B = 180^\circ - 27^\circ12'30'' = \boxed{152^\circ47'30''}$$ $$\angle D = 180^\circ - 118^\circ23'04'' = \boxed{61^\circ36'56''}$$ As a check, $\angle B$ equals the angle from $\text{Az}_{BA}=241^\circ24'10''$ to $\text{Az}_{BC}=88^\circ36'40''$, i.e. $152^\circ47'30''$; and $\angle D$ from $\text{Az}_{DC}=350^\circ26'28''$ to $\text{Az}_{DE}=288^\circ49'32''$, i.e. $61^\circ36'56''$ — both agree.

Station	Deflection angle	Interior angle
B	$27^\circ12'30''$ R	$152^\circ47'30''$
C	$81^\circ49'48''$ R	—
D	$118^\circ23'04''$ R	$61^\circ36'56''$

Concept & Theory

Topic. Relating line bearings/azimuths to deflection angles and interior angles along a route traverse.

Key relations. Bearing → azimuth by quadrant; deflection $\delta = \text{Az}_{\text{fwd}} - \text{Az}_{\text{back}}$ (right if $+$, left if $-$, reduced to $\pm180^\circ$); interior angle $=180^\circ - \delta$; back azimuth $=$ forward azimuth $\pm180^\circ$.

Why this works. The deflection angle is simply how much the direction of travel turns at a station, which is the difference of the two line azimuths. The interior angle is the supplement of the deflection because the incoming line, extended straight, and the reversed incoming line differ by exactly $180^\circ$. Working in azimuths keeps the signs automatic and avoids quadrant confusion.

Common pitfall. Forgetting to convert bearings to azimuths first (mixing $N/S$ references), or reducing a difference outside $\pm180^\circ$ incorrectly. Track the R/L sense so the supplement gives the correct interior side.

Source. Wolf & Ghilani, Elementary Surveying §7 (bearings, azimuths, deflection angles) and §9 (angle observations).

Question 7 of 7: Deflection Angles and Interior Angles from Bearings

Question 7: Deflection Angles and Interior Angles from Bearings (20 marks)