Latitude and Departure

Lat = L cos α Dep = L sin α

Latitude is the northing component (N positive); departure is the easting (E positive). α is the azimuth from north. For bearings, apply the quadrant signs: NE (+Lat, +Dep), SE (−Lat, +Dep), SW (−Lat, −Dep), NW (+Lat, −Dep).

$L$: Horizontal length of the course
$α$: Azimuth from north
$Lat$: Latitude (ΔN)
$Dep$: Departure (ΔE)

Step-by-step solved example

Course 250.00 ft, azimuth 30°00'. Find latitude and departure.

Lat = L cosα, Dep = L sinα on each course.

1. Latitude
Lat = 250 cos 30° = 250 × 0.866025 = 216.51 ft N.
$Lat = 216.51 ft N$
2. Departure
Dep = 250 sin 30° = 250 × 0.500 = 125.00 ft E.
$Dep = 125.00 ft E$

Answer: Lat = 216.51 ft N, Dep = 125.00 ft E

10 practice questions

0/10 correct

1.L = 400 ft, azimuth 90°. Lat, Dep =

2.A due-north course has departure

3.L = 100 m, azimuth 240°. Lat is nearest

4.Bearing S 30° E, L = 200 ft. Departure is

5.Latitude of a course with azimuth 180° and L = 85.00 m is

6.If Lat = +80 ft and Dep = −60 ft, length L is

7.NW quadrant signs are

8.Azimuth from Lat = +50, Dep = +50 (ft) is

9.Doubling L at the same azimuth

10.L = 500.00 ft, bearing N 53.13° E. Lat, Dep nearest