'westward the perpendicular CB, which will represent the parallel of latitude, and on it set off CB=79 miles, the departure. Join AB.

The angle A is the course, and the line AB is the distance.

Computation.

Difference of latitude AC—78 1.89209
: Radius

10'00000
:: Departure CB=79

1.89763

5. A ship in latitude 30° 5' N. sails SE6E. till her departure is 46 miles. What is her present latitude, and what distance has she run ?

Construction.

Draw the meridian NS, and from any point, A, in the same, draw the parallel of latitude Ab to the eastward = 46 miles, or the easting. From A draw AB in the SE. quarter, unlimited toward B ; and making the angle BAC

= 56° 15', or the course. From b draw bB parallel to AC, and intersecting AB in B, Draw likewise BC parallel to bA, and intersecting AC in C.

The

The line AB will be the distance, and AC the difference of latitude.

Sin. Co. ZA=56° 15' ar. co. Oʻ08015 :: Departure CB=46

1'66276 :: Cos. Co. 33° 45'

974474

: Difference of latitude=3007

1'48765

Now the difference of latitude is southing, and therefore of a different name from the given latitude. Their difference, or 30° 5'31'=29° 34', is the present latitude, which is north, because the greater of the two latitudes is north.

6. A ship sails on a southerly course 118 miles, and makes 83 miles westing. Her course and difference of latitude are required.

Construction.

N

الا

Draw the meridian NS, and from any point A in it draw the parallel of latitude Ab, to the westward, am 33 miles. From 6 draw bB parallel to AC, and unlimited toward B. With the extent 118, or the dis

B tance, from A describe an arc cutting bB in B. Join AB, and draw. BC parallel to bA, intersecting the meridian in C.

The angle BAC will be the course, and AC the differcuice of latitude.

: Difference of latitude 838

B

7. A ship from latitude 1° 4' S. sails between the N. and W., 162 miles, and then finds her latitude by observation to be 52' N. It is required to find her course and departure.

Construction, Draw the meridian

N NS, and upon it set off AC-1° 56', or 116 miles, the difference of latitude. From C, the latitude come to, draw the parallel of latitude CB unlimited to the westward. From A, the latitude sailed from, with the extent AB 162, the distance, describe an arc intersecting the parallel in B; and join AB.

The angle BAC will be the course, and CB the departure.

Computation,
Distance AB=162

220951
: Radius

10'00000 :: Difference of latitude AC=116 2'06456 : Cosine Course 45° 45'

9*85505

Radius
: Distance AB=162
:: Sine Course 44° 15'
; Departure 113

8. There are two islands, A and B, and the island B bears WNW. of a. Now a ship, after running 30 miles due west from A, observed the island B to bear N. What is the distance of the two islands ?

Ans. 32'5. miles, or 11 leagues, nearly. 9. A frigate sails on a due south course 62 miles from a port in latitude 180 so' N. at which time she speaks with a merchantman, who had observed a privateer of the enemy cruizing in the same latitude, viz. 18° 10' N. The merchantman's course thence was ESE. How far distant is the privateer, supposing her not to have changed her station ?

Ans, 162 miles.

10. A ship in sight of Cape St. Vincents bearing WbS. distant 9 leagues, finds her latitude by observation 36° 46'

9 N. The latitude of the cape is required.

Ans. 36° 41' N

Though almost every thing, which occurs in the prac: tice of navigation, may be solved by the use and application of right-angled triangles ; yet there are many instances, especially in coasting or surveying harbours, which are much more readily and elegantly solved by the help of oblique-angled triangles, and some, which can be solved no other

way. This method of solution is usually termed OBLIQUE SAILING.

EXAMPLES

1. A ship sailing on a NW. course, at the rate of 5 knots an hours observes a head land bearing E. and four