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

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BC parallel to bA, intersecting the meridian in C. The angle BAC will be the course, and AC the difference of latitude.

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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.

Draw the meridian NS, and upon it set off AC1° 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,

Construction,

B

The angle BAC will be the course, and CB the de

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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. 325 miles, or 11 leagues, nearly.

9. A frigate sails on a due south course 62 miles from a port in latitude 18° 10' 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.

TO. A ship in sight of Cape St. Vincents bearing WbS. distant 9 leagues, finds her latitude by observation 36° 46' N. The latitude of the cape is required.

Ans. 36° 41 N.

OBLIQUE PLANE SAILING.

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 other way. This method of solution is usually termed

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OBLIQUE SAILING.

EXAMPLES.

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

hours

hours afterward the same head land bore S. 78° E. What was its distance at the last time it was set?

Construction.

W

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With the radius or chord of 60°, describe the circle NESW, representing the horizon; through the centre of which draw the meridian NS, and the parallel of latitude WE. From A draw AB 20 miles on a NW. course, because the 5 knots or miles per hour, for 4 hours, make that quantity. Draw AC on an E. course, unlimited toward C. Draw AD likewise on a S. 78° E. course, by setting off from S the arc SD 78°. And parallel to AD, through B, draw BC, which will make the same angles with NS toward the same part, and consequently will on the same rhumb as AD. Prolong BC till it meet AC in C.

be

The

The side BC of the triangle ABC is the distance re

quired.

Computation.

In the triangle ABC, the angle BAC

BAN+

ZNAE 45°+90°135°. And the angle ACB is equal to the difference of the bearings or positions of the lines CA and CB from the point C; that is, between W. and N. 78 W. or 90°-78° 12°. Therefore,

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2. Running on a NE E. course at the rate of 6 knots, at II A. M.* the east point of Praya Bay, on the island of St. Jago, bore N. and the westermost point, or Point Tubaron, bore NW. At noon both points in onet bore S. 84° W. It is required to determine how wide the entrance of the bay is.

Construction.

* A. M. signifies Ante Meridiem, or before noon; P. M. Post Meridiem, or afternoon.

letters as they stand.

Seamen read them by naming the

+ When two objects, as, for example, pilots' marks, points of land, &c. are seen on the same rhumb or bearing, so that the remoter is immediately behind the nearer object, they are said to be in one.

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