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TEC-GIS45-01 · Lesson 4 · 24 clock hours · theory

Trade Math II: Geometry, Trigonometry, and Area/Volume

Objective. Students apply plane geometry and right-triangle trigonometry (including law of sines and law of cosines) to solve bearing, traverse, area, and earthwork volume problems representative of surveying field computations.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by an industry-certified professional or by a certification or approval body. It is a study draft and may contain errors. Do not use it as instructions for work on a job site or on live equipment. Always follow your instructor, your supervising professional, and current official references and regulations.

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Why Geometry and Trigonometry Matter in Surveying

Every measurement you record in the field becomes a number in a calculation. If the math is wrong, the map is wrong, the property line is wrong, and the earthwork bid is wrong. Trade Math II builds the geometric and trigonometric toolkit you will use daily as a survey technician, GIS analyst, or drone operator interpreting photogrammetric output. Accuracy is not optional: a 0.01-foot error in closure can fail a legal description, and a volume error in a cut/fill estimate can cost a contractor thousands of dollars.

Plane Geometry Fundamentals

A traverse is a series of connected lines (called courses) whose endpoints are survey stations. The interior angles of any closed polygon must satisfy a simple rule: the sum of interior angles equals (n − 2) × 180°, where n is the number of sides. For a four-sided traverse, the sum must be exactly 360°. When your measured angles do not sum to that value, the difference is the angular misclosure, which must be distributed (adjusted) among the measured angles before you can compute coordinates.

Bearings and azimuths both describe the direction of a course. A bearing is expressed as an angle from north or south toward east or west (e.g., N 42° 15′ E). An azimuth is a clockwise angle measured from north, ranging from 0° to 360°. Converting between them is a routine task: a bearing of S 42° 15′ W equals an azimuth of 180° + 42° 15′ = 222° 15′.

The Coordinate Method for Traverse Closure

Once angles are adjusted, you compute latitudes (north–south components) and departures (east–west components) for each course:

For a perfectly closed traverse the sum of all latitudes equals zero and the sum of all departures equals zero. Any remaining sum is the linear misclosure, and the ratio of misclosure to total perimeter is the precision ratio (e.g., 1:5,000). The compass rule (Bowditch adjustment) distributes the misclosure proportionally to each course length. Once adjusted coordinates are known, the coordinate method (also called the double-meridian distance or shoelace formula) gives the enclosed area precisely.

Right-Triangle Trigonometry: SOH-CAH-TOA

For any right triangle with an acute angle θ:

A common field application is slope correction: if you tape a distance along ground that slopes at angle α, the horizontal distance = slope distance × cos α. On a 5% grade, α = arctan(0.05) ≈ 2° 52′, so the correction is small but accumulates over long distances.

Law of Sines and Law of Cosines

When triangles are not right triangles, two additional rules apply.

Law of Sines: a / sin A = b / sin B = c / sin C

Use this when you know two angles and one side (AAS or ASA), or two sides and the angle opposite one of them (SSA — use carefully; the ambiguous case can produce two solutions).

Law of Cosines: c² = a² + b² − 2ab cos C

Use this when you know three sides (SSS) or two sides and the included angle (SAS). A classic surveying use is computing an inaccessible baseline across a pond: you measure two accessible sides of a triangle and the included angle, then solve for the third side.

Average-End-Area Method for Earthwork Volumes

When a road or grading project is divided into cross-sections at regular intervals, the volume of material between two adjacent sections is:

V = L × (A₁ + A₂) / 2

where L is the distance between sections, A₁ and A₂ are the cross-sectional areas (in square feet or square meters), and V is volume (in cubic feet or cubic meters, then converted to cubic yards by dividing by 27). The average-end-area method slightly overestimates volume for converging sections; the prismatoid formula is more precise, but average-end-area is acceptable for many earthwork estimates and is the standard starting point for the CST exam.

Safety in the Field and at the Desk

Math errors carry real safety consequences. An incorrect volume calculation can lead to under-designed embankments that fail. When working outdoors on traverse fieldwork:

At the computer, save field-book data and spreadsheet files regularly and maintain backups; data loss in professional practice is a serious liability issue.

Field-Book Discipline

All intermediate steps — angle sums, latitude/departure tables, adjustment corrections, area calculations — must appear in your field book in ink, with no erasures (cross out errors with a single line and initial). Entries should include the date, project name, instrument type, and your name. Professional surveyors are required by most state boards to maintain legible, complete records; the CST exam rewards this discipline.

Key terms10

TraverseA series of connected survey lines (courses) forming an open or closed polygon, used to determine the positions of points on the ground.
BearingA direction expressed as an angle between 0° and 90° measured from north or south toward east or west; written as N 42° 15′ E, for example.
AzimuthA horizontal direction measured clockwise from north, ranging from 0° to 360°, commonly used in surveying and drone navigation.
Latitude (traverse)The north–south component of a traverse course, equal to the course length multiplied by the cosine of the bearing angle.
DepartureThe east–west component of a traverse course, equal to the course length multiplied by the sine of the bearing angle.
Angular MisclosureThe difference between the measured sum of interior angles of a closed traverse and the geometric requirement of (n − 2) × 180°.
Precision RatioThe ratio of the linear misclosure of a closed traverse to its total perimeter, expressed as 1:X; a higher X indicates greater precision.
Law of CosinesThe rule c² = a² + b² − 2ab cos C, used to find an unknown side or angle in any triangle when the other two sides and included angle, or all three sides, are known.
Average-End-Area MethodA volume estimation technique that averages the cross-sectional areas of two adjacent stations and multiplies by the distance between them: V = L × (A₁ + A₂) / 2.
Slope CorrectionThe adjustment applied to a measured slope distance to obtain the shorter horizontal distance; horizontal distance = slope distance × cos(slope angle).

Worked examples4

1. Bearing-to-Azimuth Conversion: Convert the bearing S 67° 30′ W to an equivalent azimuth.
  1. Identify the quadrant. S … W means the direction is in the southwest quadrant, which spans azimuths from 180° to 270°.
  2. For a south-based bearing, azimuth = 180° + the bearing angle. The bearing angle is 67° 30′.
  3. Azimuth = 180° 00′ + 67° 30′ = 247° 30′.
  4. Verify: 247° 30′ falls between 180° and 270° (SW quadrant). ✓
Answer: Azimuth = 247° 30′
2. Slope Correction: A surveyor tapes a distance of 312.45 ft along ground that slopes at a uniform 5% grade. What is the horizontal distance?
  1. Convert the percent grade to an angle. Grade = rise / run = 0.05, so slope angle α = arctan(0.05) = 2° 51′ 45″ (approximately 2.8629°).
  2. Apply the slope-correction formula: horizontal distance = slope distance × cos α.
  3. cos(2.8629°) = 0.99875 (use your calculator; keep at least five decimal places).
  4. Horizontal distance = 312.45 ft × 0.99875 = 312.06 ft (rounded to the nearest 0.01 ft).
  5. Quick check: the correction is 312.45 − 312.06 = 0.39 ft, which is reasonable for a 5% grade over about 312 ft.
Answer: Horizontal distance = 312.06 ft
3. Inaccessible Baseline (Law of Cosines): A pond blocks a direct measurement. From instrument station A, surveyors measure AB = 185.0 ft and AC = 210.0 ft, with the angle at A between the two lines (∠BAC) = 48° 20′. Find the distance BC across the pond.
  1. Label the triangle: a = BC (unknown), b = AC = 210.0 ft, c = AB = 185.0 ft, and the included angle at A = 48° 20′ = 48.3333°.
  2. Apply the Law of Cosines: a² = b² + c² − 2bc cos A.
  3. Compute each term: b² = 210.0² = 44,100.00; c² = 185.0² = 34,225.00; 2bc = 2 × 210.0 × 185.0 = 77,700.00; cos(48.3333°) = 0.66446.
  4. 2bc cos A = 77,700.00 × 0.66446 = 51,624.42.
  5. a² = 44,100.00 + 34,225.00 − 51,624.42 = 26,700.58.
  6. a = √26,700.58 = 163.40 ft (rounded to 0.01 ft).
Answer: BC ≈ 163.40 ft
4. Average-End-Area Volume: Two adjacent cross-sections of a road cut are 50.0 ft apart. Section 1 has a cut area of 124.6 sq ft and Section 2 has a cut area of 98.2 sq ft. Find the volume of excavation in cubic yards.
  1. Apply the average-end-area formula: V (ft³) = L × (A₁ + A₂) / 2.
  2. Sum the areas: 124.6 + 98.2 = 222.8 sq ft.
  3. Average area: 222.8 / 2 = 111.4 sq ft.
  4. Volume in cubic feet: 50.0 ft × 111.4 sq ft = 5,570.0 cu ft.
  5. Convert to cubic yards (1 cu yd = 27 cu ft): 5,570.0 / 27 = 206.30 cu yd (rounded to 0.01 cu yd).
Answer: Volume ≈ 206.30 cubic yards

The math in this lesson3

Learners compute areas and volumes for survey parcels and earthwork alongside bearing and triangle solutions.
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Brush up in the math halls at newSchoolWorks.com:
In class: Solve a 15-problem set covering interior angles of traverses, bearing-to-azimuth conversion, and closure of a four-sided polygon traverse using the coordinate method; show all work in the field book and verify the answer by plotting the traverse at scale on graph paper. · Use SOH-CAH-TOA and the law of sines/cosines to compute unknown distances and angles in six surveying diagrams (e.g., inaccessible baseline across a pond, slope correction for taped distance on a 5% grade); pair-check answers and explain any discrepancy exceeding 0.01 ft. · Calculate earthwork cut/fill volumes for a simple grading plan using the average-end-area method; enter measurements into a spreadsheet template, verify the formula output manually, and present the result in a memo formatted to professional standards.
Materials: Scientific calculators; graph paper; pencils/straight edges; field books; spreadsheet application (Microsoft Excel or LibreOffice Calc); printed traverse and grading diagrams; NSPS CST Study Guide (traverse and computation chapters)
Check: 25-problem written exam covering geometry, trig, traverse closure, and volume; passing score 80%; all intermediate work shown in field book; spreadsheet memo reviewed for correct formula use and professional formatting.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

NSPS CST L1-2 — Traverse Computation
1. The sum of interior angles for a closed five-sided traverse must equal:
NSPS CST L1-2 — Directions and Bearings
2. A bearing of N 35° 45′ W converted to an azimuth equals:
NSPS CST L1-2 — Traverse Closure
3. In a closed traverse, the sum of all latitudes is +0.08 ft and the sum of all departures is −0.06 ft. The linear misclosure is:
NSPS CST L1-2 — Distance Measurement and Corrections
4. A slope distance of 500.00 ft is measured on a 6% grade. Using the cosine of the slope angle, which of the following is the best approximation of the horizontal distance?
NSPS CST L1-2 — Triangles and Trigonometry
5. You know two sides of a triangle (a = 90 ft, b = 75 ft) and the included angle C = 55°. Which formula correctly begins the solution for the third side c?
NSPS CST L1-2 — Earthwork and Volumes
6. Using the average-end-area method, two cross-sections 40 ft apart have areas of 80 sq ft and 120 sq ft. The volume in cubic yards is closest to:
NSPS CST L1-2 — Traverse Closure and Precision
7. Which of the following best describes a precision ratio of 1:10,000 for a closed traverse?
NSPS CST L1-2 — Triangles and Trigonometry
8. A surveyor uses the Law of Sines to find an unknown angle in a triangle and gets two mathematically valid solutions. This situation is known as:
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