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ADmit a Ship is at A, in Latitude 51 deg. 22 min. North, as is Lundy, and sails, or is to sail to E, in Latitude 13 deg. 10 min. according to the Plain Chart corrected, which is Barbadoes; or by Mercator's Chart, Barbadoes is in the Point D, and the Difference of Longitude is 1058 Leagues, which is 52 deg. 55 min.. First find the Difference of Latitude inlarged, as is before-directed in the first Problem, and found to be 934 5/10 Leagues.
Now you have given AB the Difference of Latitude 38 deg. 12 min. inlarged from B to C, and CD the Difference of Longitude 52 deg. 55 min. whereby the Angles and Hypothenusal shall be found by the Fourth and Fifth Case of Plain Triangles.
But because in this kind of Projection, the Degrees of Longitude and Latitude are not equal (except in Places near the Aequinoctial) the Degrees of Latitude at every Parallel, exceeding the Degrees of Longitude, in such proportion as the Aequinoctial exceeds that Parallel; therefore these Differences of Longitude and Latitude must be expressed by some one common measure; and for that purpose serves the foregoing Table, which sheweth how many Equal Parts are from the Aequinoctial, in every Degree of Latitude, to the Poles; namely, of such Equal Parts as a Degree of Lon∣gitude contains 20 Leagues.
Wherefore, as before-directed, multiplying 52 deg. by 20, and dividing the odd Minutes, being 55, by 3, it will be 18 ⅓ Leagues; added to the former Sum, makes 1058 ⅓ Leagues, for the Meridional parts contained in the Difference of Longitude. Also by the last Problem, I find the Meridional parts contained in the Difference of Latitude to be 934 5/10 Leagues: So that AC is 934 5/10 Parts, and CD 1058 ⅓ of such Parts.