Technical Geometry, Trigonometry, and Surveying

Technical geometry turns measurements into dependable positions, lengths, areas, and quantities. It is used in construction, mapping, drainage, machining, forestry, and site work. You should already be comfortable with right-triangle trigonometry, the Pythagorean theorem, and coordinate geometry; the related lessons on Trigonometry and Coordinate Geometry are useful preparation. In field work, keep units consistent, state the precision of measurements, and distinguish a measured value from a calculated one.

Directions: Bearings and Coordinate Components

A bearing is an angle measured clockwise from north and written with three digits: 065°, 180°, and 305°. On a local coordinate plan, take east as positive x and north as positive y. For a line of horizontal length d on bearing θ, its coordinate changes are:

ΔE = d sin θ and ΔN = d cos θ.

The sine and cosine are deliberately reversed from the familiar “angle from the positive x-axis” convention, because a bearing starts at north. A negative component means west or south.

Worked example: locating a point from a bearing

A survey station is 240 m from a benchmark on bearing 065°. Find its east and north offsets.

ΔE = 240 sin 65° = 217.5 m
ΔN = 240 cos 65° = 101.4 m

The station is 217.5 m east and 101.4 m north
of the benchmark.

Use a calculator in degree mode unless the instrument or problem explicitly gives radians.

Worked example: bearing from coordinate differences

A point is 84 m east and 112 m north of a station. Its distance is √(842 + 1122) = 140 m. Since it lies northeast, its bearing is measured east of north:

θ = tan⁻¹(84 / 112) = 36.9°

Bearing = 037° (to the nearest degree)

Use the signs of both coordinate differences to select the quadrant. An inverse tangent alone does not tell you whether the direction is northeast, southeast, southwest, or northwest.

Elevation, Depression, and Slope

An angle of elevation is measured upward from a horizontal line of sight. An angle of depression is measured downward from horizontal. Horizontal lines at two different heights are parallel, so the angle of depression from an observer equals the corresponding angle of elevation from the object.

Worked example: height from an angle of elevation

An instrument is 1.55 m above level ground and is 82.0 m horizontally from a tower. The angle of elevation to the top is 36.0°. Find the tower height.

vertical rise above instrument = 82.0 tan 36.0° = 59.58 m
tower height = 59.58 + 1.55 = 61.13 m

To suitable precision, the tower is 61.1 m high.

The 82.0 m is the horizontal distance, not the sloping line-of-sight distance. If the ground between observer and tower is not level, use elevations or a properly reduced horizontal distance.

Grade, ratio, and angle

A slope compares vertical rise to horizontal run. Its grade is usually a percentage:

grade = (rise/run) × 100%, while tan α = rise/run.

A slope of 1 in 12 rises 1 unit for every 12 horizontal units. Therefore its grade is (1/12) × 100% = 8.33%, and its angle is tan−1(1/12) = 4.76°. “Percent grade” is not the same number as degrees.

Finding an Inaccessible Distance

When an object cannot be reached, measure a known baseline between two accessible points and angles to the object. The sine rule then solves the triangle. For a triangle with sides a, b, c opposite angles A, B, C:

a/sin A = b/sin B = c/sin C.

Worked example: distance across a river

Points A and B on one bank are 120 m apart. A tree C is across the river. The interior angle at A is 52° and the interior angle at B is 71°. Find AC.

∠C = 180° − 52° − 71° = 57°

AC / sin 71° = 120 / sin 57°
AC = 120 sin 71° / sin 57°
AC = 135.4 m

The angles must be the interior angles of the same triangle. If observations are recorded as bearings, first convert them carefully into the angle between the sight lines.

Areas from Dimensions and Coordinates

Break a practical shape into rectangles, triangles, circles, or sectors when that is clearest. For a closed polygon with coordinate vertices (x1, y1), …, (xn, yn), the shoelace formula gives:

A = 1/2 |Σ(xiyi+1) − Σ(yixi+1)|, where the first vertex is repeated at the end.

Worked example: a parcel from coordinates

A parcel has vertices, in order, (0, 0), (80, 0), (80, 35), (50, 50), (0, 35), in metres.

Σ(xᵢyᵢ₊₁) = 0 + 2800 + 4000 + 1750 + 0 = 8550
Σ(yᵢxᵢ₊₁) = 0 + 0 + 1750 + 0 + 0 = 1750

Area = 1/2 |8550 − 1750|
     = 3400 m²

List boundary points consecutively around the parcel. Crossing from one side to another while listing points can produce an incorrect area.

Volume and Surface Area for Materials

Volume measures space and is expressed in cubic units; surface area measures covering and uses square units. For a cylinder of radius r and height h, V = πr2h. Its total outside area, including both circular ends, is 2πr2 + 2πrh.

Worked example: a cylindrical tank

A closed cylindrical tank has radius 1.20 m and height 3.50 m.

Volume = π(1.20)²(3.50) = 15.83 m³

Total surface area = 2π(1.20)² + 2π(1.20)(3.50)
                   = 35.44 m²

For paint, decide whether the base, top, or inside is actually included. For capacity, convert only after calculating: 15.83 m3 is about 15 830 L.

Traverse Closure and Field Checks

A closed traverse starts at a known point, follows measured lines, and returns to that point. Ideally, the sums of all easting and northing changes are both zero. In reality, small observation errors create a closure error.

Worked example: calculate a simple closure

A rectangular-looking traverse records 100.0 m north, 75.0 m east, 100.6 m south, and 74.2 m west.

ΣΔE = 75.0 − 74.2 = +0.8 m
ΣΔN = 100.0 − 100.6 = −0.6 m

linear misclosure = √[(0.8)² + (−0.6)²] = 1.0 m
perimeter = 349.8 m
relative precision = 349.8 / 1.0 ≈ 1 : 350

The computed endpoint is 0.8 m east and 0.6 m south of the start. A professional survey follows its required standards and may distribute an acceptable small misclosure by a stated method, often in proportion to line lengths. Never silently “force” coordinates to close.

Common Mistakes

  • Measuring a bearing counterclockwise from east instead of clockwise from north.
  • Using a sloping distance as the adjacent side in a tangent calculation.
  • Forgetting instrument height or target height in an elevation calculation.
  • Mixing metres with millimetres, or reporting an area in metres instead of square metres.
  • Rounding each coordinate component early; retain guard digits and round the final result sensibly.
  • Treating a closure error as proof that one particular line is wrong. Closure reveals combined error, not its source.

Practice Set

  1. A line is 180 m on bearing 128°. Find its easting and northing components to the nearest tenth of a metre.
  2. From a station, a point is 45 m west and 60 m south. Find the distance and whole-degree bearing.
  3. A 2.0 m-high instrument is 48 m horizontally from a roof. Its angle of elevation to the roof is 28°. Find the roof height above ground.
  4. A ramp rises 0.75 m over a horizontal run of 9.0 m. Find its grade and its angle to the horizontal.
  5. A 90 m baseline has angles 48° and 67° to an inaccessible point. Find the distance from the 48° end to the point.
  6. Use the shoelace formula to find the area of the triangular lot with vertices (10, 5), (70, 5), and (40, 45), in metres.
  7. Find the volume and total surface area of a closed cylinder with radius 0.50 m and height 2.0 m. Give exact answers in terms of π and decimal approximations.
  8. A traverse has total coordinate changes ΣΔE = −0.30 m and ΣΔN = +0.40 m over a 500 m perimeter. Find the linear misclosure and relative precision.

Answer Checks

  1. ΔE = 180 sin 128° = 141.8 m; ΔN = 180 cos 128° = −110.8 m. The negative northing component means 110.8 m south.
  2. Distance = √(452 + 602) = 75 m. The point is southwest. It is 36.9° west of south, so the bearing is 217°.
  3. Height = 2.0 + 48 tan 28° = 27.5 m, to the nearest tenth.
  4. Grade = (0.75/9.0) × 100% = 8.33%. Angle = tan−1(0.75/9.0) = 4.76°.
  5. The third angle is 180° − 48° − 67° = 65°. The requested distance is 90 sin 67°/sin 65° = 91.5 m.
  6. A = 1/2 |(10·5 + 70·45 + 40·5) − (5·70 + 5·40 + 45·10)| = 1200 m2.
  7. V = π(0.50)2(2.0) = 0.5π m3 ≈ 1.57 m3. Surface area = 2π(0.50)2 + 2π(0.50)(2.0) = 2.5π m2 ≈ 7.85 m2.
  8. Misclosure = √[(0.30)2 + (0.40)2] = 0.50 m. Relative precision = 500/0.50 = 1 : 1000.