Latitudes and Departures

Surveying · FE Reference Handbook section

Surveying

0 formulas

10 exam-style examples

~45 min

All Surveying lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Latitudes and Departures within Surveying. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what latitudes and departures describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: bearings in quadrant form, azimuths from north.

Lecture

Why this section exists. Latitudes and Departures is the part of Surveying that lets you connect a closed traverse or a cut-and-fill section to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a closure, an area, or an earthwork volume. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. bearings in quadrant form, azimuths from north. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Surveyor operating a total station beside a rural highway with a rod person downstream. — Photo 1. Where this shows up in practice: latitudes and departures.
Capstone Studio instructional photograph

Surveying — Latitudes and Departures: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a closed traverse or a cut-and-fill section. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

+ Latitude
– Departure + Departure
– Latitude

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Traverse closure and precision

A closed traverse has ΣLatitudes = +0.084 m and ΣDepartures = −0.112 m over a total length of 1,240 m. Find the linear misclosure and the precision.

Given

ΣLat = +0.084 m
ΣDep = −0.112 m
Perimeter = 1,240 m

Find

Misclosure and precision ratio

Start with the thinking

Misclosure is the vector sum of the two errors.
Precision is expressed as 1 in N.

Step-by-step solution

Misclosure
$e = (0.08 4^{2} + 0.11 2^{2}) = (0.00706 + 0.01254)$
Evaluate
$e = 0 .01960 = 0.140 m$
Precision
$P = e / p er im e t er = 0.140/1, 240 = 1.13 \times 1 0^{- 4}$
Express
$1 : 1/1.13 \times 1 0^{- 4} = 1 : 8, 860$

Answer: e = 0.140 m; precision ≈ 1:8,900

Why the other options are there

e = 0.196 m (errors added directly)
1:1,240 (perimeter reported as precision)

Reference: FE Reference Handbook — Surveying — Traverse computations

Example 2

Differential leveling elevation

A benchmark at elevation 214.560 m is sighted with a backsight of 1.842 m. The foresight to the new point reads 2.615 m. What is the new elevation?

Given

BM = 214.560 m
BS = 1.842 m
FS = 2.615 m

Find

Elevation of the new point

Start with the thinking

Height of instrument = elevation + backsight.
Foresight is subtracted from HI.

Step-by-step solution

Height of instrument
$H I = 214.560 + 1.842 = 216.402 m$
New elevation
$E l e v = H I - F S = 216.402 - 2.615$
Result
$E l e v = 213.787 m$
Check — the point is 0.773 m below the benchmark, matching FS − BS ✓

Answer: Elevation = 213.787 m

Why the other options are there

215.333 m (BS and FS reversed)
219.017 m (FS added)

Reference: FE Reference Handbook — Surveying — Differential leveling

Example 3

Latitude and departure of a traverse course — Latitudes and Departures

A traverse course is 466 ft long with an azimuth of 41° from north. Compute its latitude and departure.

Given

L = 466 ft
Azimuth = 41°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 466·cos 41° = 351.7 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 466·sin 41° = 305.7 ft
Check
$(L a t^{2} + D e p^{2}) = 466.0 f t = L ✓$

Answer: Lat = 351.7 ft, Dep = 305.7 ft

Why the other options are there

Lat = 305.7 ft (sine and cosine swapped)
Lat = -460.1 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Example 4

Latitude and departure of a traverse course — Latitudes and Departures (2)

A traverse course is 357 ft long with an azimuth of 48° from north. Compute its latitude and departure.

Given

L = 357 ft
Azimuth = 48°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 357·cos 48° = 238.9 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 357·sin 48° = 265.3 ft
Check
$(L a t^{2} + D e p^{2}) = 357.0 f t = L ✓$

Answer: Lat = 238.9 ft, Dep = 265.3 ft

Why the other options are there

Lat = 265.3 ft (sine and cosine swapped)
Lat = -228.5 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Example 5

Latitude and departure of a traverse course — Latitudes and Departures (3)

A traverse course is 341 ft long with an azimuth of 264° from north. Compute its latitude and departure.

Given

L = 341 ft
Azimuth = 264°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 341·cos 264° = -35.64 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 341·sin 264° = -339.1 ft
Check
$(L a t^{2} + D e p^{2}) = 341.0 f t = L ✓$

Answer: Lat = -35.64 ft, Dep = -339.1 ft

Why the other options are there

Lat = -339.1 ft (sine and cosine swapped)
Lat = 339.1 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Example 6

Latitude and departure of a traverse course — Latitudes and Departures (4)

A traverse course is 443 ft long with an azimuth of 319° from north. Compute its latitude and departure.

Given

L = 443 ft
Azimuth = 319°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 443·cos 319° = 334.3 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 443·sin 319° = -290.6 ft
Check
$(L a t^{2} + D e p^{2}) = 443.0 f t = L ✓$

Answer: Lat = 334.3 ft, Dep = -290.6 ft

Why the other options are there

Lat = -290.6 ft (sine and cosine swapped)
Lat = 56.70 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Example 7

Latitude and departure of a traverse course — Latitudes and Departures (5)

A traverse course is 540 ft long with an azimuth of 188° from north. Compute its latitude and departure.

Given

L = 540 ft
Azimuth = 188°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 540·cos 188° = -534.7 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 540·sin 188° = -75.15 ft
Check
$(L a t^{2} + D e p^{2}) = 540.0 f t = L ✓$

Answer: Lat = -534.7 ft, Dep = -75.15 ft

Why the other options are there

Lat = -75.15 ft (sine and cosine swapped)
Lat = 475.0 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Example 8

Latitude and departure of a traverse course — Latitudes and Departures (6)

A traverse course is 530 ft long with an azimuth of 124° from north. Compute its latitude and departure.

Given

L = 530 ft
Azimuth = 124°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 530·cos 124° = -296.4 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 530·sin 124° = 439.4 ft
Check
$(L a t^{2} + D e p^{2}) = 530.0 f t = L ✓$

Answer: Lat = -296.4 ft, Dep = 439.4 ft

Why the other options are there

Lat = 439.4 ft (sine and cosine swapped)
Lat = -49.17 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Example 9

Latitude and departure of a traverse course — Latitudes and Departures (7)

A traverse course is 566 ft long with an azimuth of 336° from north. Compute its latitude and departure.

Given

L = 566 ft
Azimuth = 336°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 566·cos 336° = 517.1 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 566·sin 336° = -230.2 ft
Check
$(L a t^{2} + D e p^{2}) = 566.0 f t = L ✓$

Answer: Lat = 517.1 ft, Dep = -230.2 ft

Why the other options are there

Lat = -230.2 ft (sine and cosine swapped)
Lat = -559.6 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Example 10

Latitude and departure of a traverse course — Latitudes and Departures (8)

A traverse course is 208 ft long with an azimuth of 28° from north. Compute its latitude and departure.

Given

L = 208 ft
Azimuth = 28°

Find

Latitude and departure

Start with the thinking

Latitude is the north–south component, departure the east–west.
Azimuths are measured clockwise from north.

Step-by-step solution

Latitude — Lat = L·cos(Az)
Substituting — Lat = 208·cos 28° = 183.7 ft
Departure — Dep = L·sin(Az)
Substituting — Dep = 208·sin 28° = 97.65 ft
Check
$(L a t^{2} + D e p^{2}) = 208.0 f t = L ✓$

Answer: Lat = 183.7 ft, Dep = 97.65 ft

Why the other options are there

Lat = 97.65 ft (sine and cosine swapped)
Lat = -200.2 ft (calculator left in radians)

Reference: FE Reference Handbook — Surveying → Latitudes and Departures

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a closed traverse or a cut-and-fill section, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Latitudes and Departures contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a closure, an area, or an earthwork volume.
Unit rule: bearings in quadrant form, azimuths from north.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

bearings in quadrant form, azimuths from north
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.