Plane trigonometry with practical applications, by Leonard E. Dickson.

62

TRIGONOMETRY

[Art. 45

We shall postpone to Part II consideration of the various corrections to the reading of the ship's compass, and assume for the
present that we know the true course.
PART I. THE SAILINGS (TRUE COURSE ASSUMED)
45. Plane Sailing. Let the distance D sailed from A to B be
so short that the curvature of the earth may be neglected. Let the
meridian AF through A and the parallel of latiF -- — B tude BF through B intersect at F (Fig. 33).
Then AF is the difference of latitude of A and B,
D A while FB is called the departure. Let angle C
be the true course. Then
Cgy ~ diff. lat. = D cos C, dep. = D sin C.
A fEXAMPLE. From latitude 36~ N, a ship sails 243 miles
FIG. 33 S 56~ W. Find lat. in (i.e., the latitude of the place
arrived at) and the departure.
Solution. Divide the distance by 3 to obtain a distance, 81, less than 100.
Multiply by 3 the entries in Traverse Table VI which are opposite to distance
81 and above the angle 56~, using the headings at the bottom of the page and
retaining only one decimal place. We get dep. = 201.4, diff. lat. = 135.9.
Dividing the latter by 60, to convert miles into degrees, we have diff. lat.
2~16'. Since the ship sailed in a southerly direction, we subtract this from
36~ N and get lat. in = 33~44' N.
In place of the first step in the solution, we may add the entries for the
distances 200 and 43 (as in Ex. 1, Art. 17).
EXERCISES ON PLANE SAILING
Find the quantities indicated by question marks, using Table VI for
Exs. 1-5, 7, and logarithms for Exs. 6, 8.

Ex. True Course Dist. Dep. Lat. out Lat. in
1. 5 188? 40~ 33' N?
2. 124~ 488? 1~ 45' N?
3. S 17 W?? 400 17' N 37~ 6' N
4.? 360? 21 59' S 24~ 49' S
5. 236? 48.2 38 N?
6.? 289.2 W 20~ 48' N 17~ 13' N