Surveying is 6–9 questions on traverse angles and distances, areas, earthwork volumes, coordinate systems and leveling. Most problems below are calculations, so keep the calculator in degree mode and carry the seconds.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Civil
questions on the exam6–9
problems here15
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What the section covers
Five topics in the July 2020 specification:
Angles, distances and trigonometry.
Area computations.
Earthwork and volume computations.
Coordinate systems: state plane, latitude and longitude.
What is the horizontal distance between two points if the slope distance is 150.0 ft and the angle of inclination is 12.5°?
Given
Slope distance S = 150.0 ft
Vertical (inclination) angle α = 12.5°
Find
Horizontal distance H (ft)
Solution
H = S cos α = 150.0 × cos 12.5° = 146.4 ft
FE Reference Handbook, Surveying — slope reduction.
32.5 ft is S·sin α, the vertical difference, not the horizontal distance.
H = S cos α = 150.0 × cos 12.5° = 146.4 ft.
150.0 ft is the slope distance with no reduction to horizontal.
153.6 ft divides by cos α; the horizontal leg of a right triangle is shorter than the slope.
Question 2 of 15 · Surveying · 6–9 items
A surveyor needs to determine the elevation of point B. The instrument height at point A is 5.2 ft, the backsight reading is 8.75 ft, and the foresight reading to point B is 4.30 ft. If the elevation of point A is 345.65 ft, what is the elevation of point B?
Given
Elevation of A = 345.65 ft
Backsight on A = 8.75 ft
Foresight on B = 4.30 ft
Instrument height above A = 5.2 ft
Find
Elevation of B (ft)
Solution
HI = ElevA + BS = 345.65 + 8.75 = 354.40 ft
ElevB = HI − FS = 354.40 − 4.30 = 350.10 ft
FE Reference Handbook, Surveying — differential leveling.
HI = 345.65 + 8.75 = 354.40 ft, so elevation B = 354.40 − 4.30 = 350.10 ft; the instrument's height above A plays no role in differential leveling.
354.40 ft is the height of instrument; the foresight was not subtracted.
355.30 ft adds the 5.2-ft instrument height, which the backsight rod reading already accounts for.
358.70 ft adds both the backsight and the foresight; the foresight must be subtracted from the height of instrument.
Question 3 of 15 · Surveying · 6–9 items
A surveyor measures the distance between points A and B as 752.36 ft. If the temperature during measurement was 85°F and the calibration temperature of the tape was 68°F, what is the corrected distance if the tape is made of steel (coefficient of thermal expansion = 6.5 × 10−6/°F)?
FE Reference Handbook, Surveying — tape corrections.
752.28 ft subtracts the correction; a hot tape expands, reads short, and the true distance is longer.
752.36 ft is the uncorrected measurement.
Ct = αL(T − Ts) = 6.5 × 10−6 × 752.36 × 17 = +0.083 ft, so the distance is 752.44 ft.
752.78 ft uses 85 °F as the temperature change instead of 85 − 68 = 17 °F.
Question 4 of 15 · Surveying · 6–9 items
A quadrilateral parcel has state plane coordinates (E, N) of (200, 300), (450, 320), (430, 600) and (180, 580), in feet. What is its area by the coordinate (shoelace) method?
FE Reference Handbook, Surveying — area by coordinates.
35,200 ft² halves the shoelace sum twice.
Σ(xiyi+1 − xi+1yi) = 140,800, and half of it is 70,400 ft².
81,000 ft² is a bounding rectangle (270 × 300 ft), not the parcel.
140,800 ft² is the shoelace sum before dividing by 2.
Question 5 of 15 · Surveying · 6–9 items
Control point X is located by intersection. From point A at (E, N) = (100, 200) the azimuth to X is 30°; from point B at (400, 200) the azimuth to X is 330°. All coordinates are in feet. What are the coordinates (E, N) of X?
FE Reference Handbook, Surveying — coordinates, azimuths.
373.2 adds the ray length 150/cos 30° = 173.2 ft as if it were a northing offset.
Northing 286.6 uses 150 tan 30°; with the azimuth measured from north, the northing offset is 150/tan 30°.
The easting and northing are swapped.
By symmetry E = 250; from A, ΔE = 150 = d sin 30°, so d = 300 ft and ΔN = 300 cos 30° = 259.8 ft, giving N = 459.8.
Question 6 of 15 · Surveying · 6–9 items
A surveyor measures a ground distance of 1,500.00 ft. The state plane grid scale factor at the site is 0.99962 and the elevation factor is 0.99985. What is the grid distance, to the nearest 0.01 ft?
FE Reference Handbook, Surveying — state plane coordinates.
Combined factor = 0.99962 × 0.99985 = 0.99947, and grid = 1,500.00 × 0.99947 = 1,499.21 ft.
1,499.43 ft applies the scale factor only and leaves out the elevation factor.
1,499.78 ft applies the elevation factor only.
1,500.00 ft is the ground distance, uncorrected.
1,500.80 ft divides by the combined factor; dividing converts grid to ground, not ground to grid.
Question 7 of 15 · Surveying · 6–9 items
Convert the quadrant bearing S 35° 12' 30" W to a full-circle azimuth. Also, for a baseline with ΔE = -57.35 m and ΔN = -81.42 m, compute the azimuth using atan2(ΔE, ΔN) and report both to the nearest 0.01°.
Az = atan2(ΔE, ΔN) = atan2(−57.35, −81.42) = 215.16°
FE Reference Handbook, Surveying — bearings and azimuths.
Both azimuths are put in the southeast quadrant (180° − angle); a W bearing adds the angle to 180°.
180° drops the 35°12′30″ angle from the bearing conversion.
144.84° = 180° − 35.16° places the line in the southeast quadrant; with both ΔE and ΔN negative it lies in the southwest.
These are the quadrant angles; a southwest bearing must be converted with Az = 180° + angle.
S 35°12′30″ W → 180° + 35.21° = 215.21°, and atan2(−57.35, −81.42) = 215.16° in the third quadrant.
Question 8 of 15 · Surveying · 6–9 items
A differential leveling run consists of 8 setups. The starting benchmark elevation is 512.48 ft. After completing the loop, the computed final elevation at the closing benchmark (known elevation 512.48 ft) is 512.35 ft. The project specification allows a maximum misclosure of K · √n where K = 0.04 ft and n is the number of setups. Compute the numeric misclosure and allowable misclosure, then determine the appropriate action.
Given
Starting benchmark elevation = 512.48 ft
Closing benchmark known elevation = 512.48 ft
Computed closing elevation = 512.35 ft
Number of setups n = 8
K = 0.04 ft
Find
Misclosure, allowable misclosure and action
Solution
e = 512.35 − 512.48 = −0.13 ft
allowable = 0.04√8 = 0.113 ft < 0.13 → re-observe
FE Reference Handbook, Surveying — leveling closure.
0.32 ft is K × n; the tolerance grows with √n, not n.
1.04 ft multiplies the misclosure by the 8 setups; the misclosure is just observed minus known.
Distributing the correction is only allowed when the misclosure is within tolerance, and 0.13 ft is not.
0.04 ft is K alone; the allowable is K√n = 0.04√8 = 0.113 ft.
Misclosure = 512.35 − 512.48 = −0.13 ft, and |0.13| > 0.04√8 = 0.113 ft, so the run fails and must be re-observed.
Question 9 of 15 · Surveying · 6–9 items
A surveyor measures an angle of 35°24'18". What is this angle expressed in decimal degrees?
Given
Angle = 35°24′18″
Find
Angle in decimal degrees
Solution
35°24′18″ = 35 + 24/60 + 18/3,600 = 35.405°
35.240° reads the minutes as a decimal (35 + 0.24); minutes must be divided by 60.
35.242° strings the minutes and seconds together as decimals (35.2418) instead of converting them.
35.400° converts the minutes (24/60 = 0.400) but drops the 18 seconds.
A property has the shape of a trapezoid with parallel sides of 250 ft and 320 ft, and a distance between them of 180 ft. What is the area of this property?
Given
Parallel sides a = 250 ft and b = 320 ft
Distance between them h = 180 ft
Find
Area (ft²)
Solution
A = ½(a + b)h = ½(250 + 320)(180) = 51,300 ft²
FE Reference Handbook, Mathematics — mensuration of areas.
45,000 ft² multiplies the shorter side by the height.
A = (a + b)h/2 = (250 + 320) × 180/2 = 51,300 ft².
57,600 ft² multiplies the longer side by the height.
102,600 ft² forgets to halve the sum of the parallel sides.
Question 11 of 15 · Surveying · 6–9 items
A highway has a grade of +3.5%. What is the elevation difference between two points on this highway that are 500 ft apart horizontally?
Given
Grade G = +3.5 %
Horizontal distance = 500 ft
Find
Elevation difference Δh (ft)
Solution
Δh = G × L = 0.035 × 500 = 17.5 ft
FE Reference Handbook, Transportation — vertical alignment (grade).
1.8 ft uses 0.0035, a decimal slip; 3.5 % is 0.035.
3.5 ft reports the grade in percent as feet, which is the rise over only 100 ft.
Δh = 0.035 × 500 ft = 17.5 ft.
175.0 ft uses 0.35 instead of 0.035.
Question 12 of 15 · Surveying · 6–9 items
A total station is set over point A and backsighted on point B; the azimuth from A to B is 135°. The clockwise angle (angle right) from the backsight to point C is 82.5°. What is the azimuth from A to C?
Given
Backsight azimuth A→B = 135°
Clockwise angle from backsight to C = 82.5°
Find
Azimuth A→C (°)
Solution
AzAC = AzAB + ∠right = 135° + 82.5° = 217.5°
FE Reference Handbook, Surveying — azimuths.
37.5° starts from the back azimuth (315°); the azimuth A→B is already measured from the instrument station.
52.5° subtracts the angle; an angle turned clockwise is added to the backsight azimuth.
135.0° is the backsight azimuth itself; the turned angle was not applied.
A land surveyor is using GPS to determine the coordinates of a point. What coordinate system directly provides latitude and longitude?
Solution
GNSS positions are computed in ECEF and reported as geodetic latitude/longitude/height; projections (SPCS, UTM) are derived from them.
A geographic system expresses position directly as latitude, longitude and ellipsoid height on a datum such as NAD 83.
A Cartesian (ECEF) system gives X, Y, Z from Earth's center, which must be converted to reach latitude and longitude.
State plane gives projected northings and eastings in feet or meters, not latitude and longitude.
UTM is a map projection with zone eastings and northings in meters.
Question 14 of 15 · Surveying · 6–9 items
A traverse has three legs: N 43°15′ E, 350 ft; S 20°30′ E, 420 ft; and S 58°45′ W, 500 ft. Starting at (E, N) = (0, 0), what are the coordinates of the terminal point?
Given
Leg 1: N 43°15′ E, 350 ft
Leg 2: S 20°30′ E, 420 ft
Leg 3: S 58°45′ W, 500 ft
Start (E, N) = (0, 0)
Find
Terminal point (E, N), ft
Solution
Lat = L cos β, Dep = L sin β (N, E positive)
ΣLat = 350 cos 43.25° − 420 cos 20.5° − 500 cos 58.75° = −397.9 ft
ΣDep = 350 sin 43.25° + 420 sin 20.5° − 500 sin 58.75° = −40.6 ft
FE Reference Handbook, Surveying — latitudes and departures.
The easting sign is lost: the west departure of leg 3 outweighs the two east departures.
Both signs are lost; departures and latitudes must carry E/W and N/S signs.
The northing sign is lost: two long south legs make the sum of latitudes negative.
ΣDep = 239.8 + 147.1 − 427.4 = −40.6 ft and ΣLat = 254.9 − 393.4 − 259.4 = −397.9 ft.
Question 15 of 15 · Surveying · 6–9 items
A civil engineering project uses control points derived from GNSS (WGS84) and existing survey monuments published in NAD83(CORS96). The project requires design-level positional accuracy. Which of the following is essential before combining these coordinates?
Solution
NAD 83 is fixed to the North American plate; WGS 84 (≈ ITRF) is not, so the offset is about 1–2 m and changes with epoch. Use NGS tools (NCAT/HTDP) for the transformation.
WGS 84 and NAD 83 differ by about 1–2 m in the conterminous US; mixing them unchecked corrupts design-level positions.
Units alone are not enough; the meter-level datum difference remains.
The two frames are not equivalent at survey accuracy.
A three-parameter average shift is too coarse for design accuracy and ignores plate motion between epochs.
Combining frames needs matching datum, realization and epoch, matching units (US survey foot vs meter) and a published transformation.
Answer every question to see your score
Common traps
Leveling signs. A backsight is added to the known elevation to get the height of instrument; a foresight is subtracted from it.
Precision of a traverse. The linear misclosure divided by the total length, written as 1 in N.
Volume formulas. Average end area overestimates the volume between unlike end sections; the prismoidal formula is exact for prismoids and needs the middle area.
Degrees, minutes, seconds. Calculator work needs decimal degrees; convert both ways without dropping the seconds.
Grid vs ground. State plane coordinates live on a projection; ground distances need the scale and elevation factors to move between the two.
Sources: FE Reference Handbook; Ghilani and Wolf, Elementary Surveying.
Bearing is not azimuth
An azimuth runs 0–360° clockwise from north; a bearing is an angle of up to 90° from north or south toward east or west. Convert every bearing to an azimuth before computing latitudes and departures, and the signs take care of themselves.
latitude = L·cos(Az) departure = L·sin(Az)
How it connects
Grades and stationing return on transportation in vertical curves; earthwork volumes feed haul and cost on construction. The trigonometry and analytic geometry behind traverses are in the common FE math section, taught on the FE practice test.
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