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Then for the Difference of Latitude P R.

As the Co-fine of the Rumb H PR, 42, 11 d.

Is to the Departure HR, 113. 89 M. So is the Sine of Rumb P HR, 47. 89 d.

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To the Side PR 126. 03 M. the Diff. of Latit.
Which reduced into deg, makes 2. 1o deg. which is the dif-
ference of Latitudes, which being added to 18. 25 deg. The
Latitude of H, from whence you came, makes 20. 35 de.
For the Latitude in which you are at P.

The Eight foregoing PROBLEMS, were all refolved by Right-angled Plain Triangles: Thefe which follow, fall under the CASES of Oblique-angled Plain Triangles.


A Ship at C, in 22 deg. of North Latitude, Sails North-Eaft Fig.

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22. 50 deg. 64. 50 Miles to D: And from D, fhe Sails LXIV. between the South and the East 124. bo Miles and then finds that she is in the fame Latitude from whence the first came: I would now know.

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I. What was the Rumb, (or Course) from D to F
II. The Difference of Longitude betwen C and F.
III. What Latitude the Ship was in, when she was at D.

The Geometrical Conftruction of the Figure.
First, Draw a Right Line NS for the Meridian; and at Right
Angles thereto another W E, for the Parallel of Latitude 22. 00
deg. North.


Secondly, Upon E, Protract the Angle of the Rumb NorthEaft 22. 50 deg. a CD and fet thereon 64.50 M. the distance fail ed from E to D; and through the point D, draw an obfcure Line Da, parallel to the Line W E.

Thirdly, Take the Distance Sailed from D to F 124.00 Miles in your Compaffes, and fetting one foot in D, with the other defcribe the Arch bc, croffing the Parallel of Latitude W E, in the point F.

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Fig. Fourthly, Joyn CD and FD, fo fhall you have conftituted LXIV. an Oblique-angled Triangle DCF, In which there is Given (1) The Side CD, 6450 M.. (2.) The Side D F 124.00 M. and (3.) The Angle DCF, 67.50 de. the Complement of the Rumb failed upon from C to D; to find (1.) The Angles at D and F, the Rumbs:. (2.) The Difference of Longitude between C and F: And (3.) The Difference of Latitude, Ca..

By Trigonometrical Calculation.

1. For the Rumb (or Course) DFC, By CASE F. of O: A. P. T..

As the Dift. failed from D to F 124. 00 M.


Is to the Co-fine of the Rumb, from C to D, 22. 50 d. ̄ 9.965615 So is the Distance failed, from C to D 64. 50 M.

1.809559 11.775174 9.681753

To the Sine of the Angle D.FC 28. 72 d..
Whofe Complement 61, 28 de. is the Rumb from F to D,
that is North-Weft 61. 28 de: And South-Eaft 61. 28 de..
from D to F.

For the Difference of Longitude C F.

The Angles at C and F being known, the Angle at D is alfo known, and will be found to be 96. 22 de. Then,

Is to the Distance failed from C to D 64. 50 M.

As the Sine of DF C, 28. 72 deg.

So is the Sine of the Angle CDF, 83.78 der






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3. For the Diff. of Latitude, DG, equal to C a.
By CASE IV of R. A. P. F.

As the Radius Sine 90 de.

To the Dift. failed from C to D, 64. 50 M..

To the Dif, of Latitude D G 3 C, 59. 59 L..



So is the Co-fine of the Rumb from C to D 67 50 d. 9.965615




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WO Ships fet Sail, from the Port P, One of them Sails
North-Weft 70. 00 deg. 106. coMiles to; The other Sails.
South-Weft 74. oo deg. 236. 60 Miles, to M. I demand,

I. How the two Ships at O and M, do Bear from each other::

H. Their Distance MQ..

The Geometrical Conftruction,

Having drawn your Meridian NS, and Parallel W E, croffing each other at Right Angles in the point P, which let be the Port from whence the two Ships failed. Then,

First, Upon P, défcribe an Arch of a Circle, as nom: And because the Courfe from P to O, was North-West 70. 00 deg. fet 20 deg. (the Complement thereof) from o tom; and through m, draw a Line, as Pm O, fetting thereon 106. oo Miles,from P to O.

Secondly, Because the Courfe from P to M, was South-west 74. · oo deg.. fet 16. oo deg. (the Complement thereof) from o to n, And through ", draw Pn, fetting thereon, the Distance failed, viz. 236. 60 Miles from P to M.

Thirdly, Joyn O and M. So fhall you have conftituted an
Oblique-Angled Triangle P O M: In which there is Given, (1.) The
Angle OPM, 36 co deg. the Sum of the Complements of the
Bearings of the two Ships. (2) The two Sides PO 106. 00 Miles
and PM 236. 60 Miles. The feveral Diftances which the two
Ships failed: And by help of thefe, you may Find. (1) The
Bearing of the two Ships one from the other: And (2.) Their
By Trigonometrical Calculation.

I: For the Bearings, O M P and MOP ;.,
By CASE II. of O. A. P. T...

The Sum of the Sides, PO and OM, is,

342.06. M.

130.06. M.

The balf Sum of the Un-known Angles M and O, is 72.00. De.

Their Difference.

Being thus prepared, fay,





As the Sum of the Sides 342. 06

LXV. Is to the Difference of the Sides 130. 06

2.534102 2.113144

So is the Tangent of the Sum of the Angles 72 d. 10.448224

12.561368 To the Tangent of 46. 80 degrees. 10.027266 Which 46. 80 deg. Added to 72. 00 deg. (the balf Sum of the Angles O and M) gives 118. 80 d. for the Angle MOP; And 46.80 d. fubftracted from 72 00 d. leaves 25. 20 deg. for the leffer Angle OMP.

Now for the Bearing, fubftract the Bearing of the Ship from P to O, viz. 72. co deg. from 118. 80 deg. (the quantity of the MOP) there will remain 48. 80 deg. for the Angle MOQ, whichis South-Weft 48. 80 d. and fo doth the Ship at O Bear from the Ship at M. And the Ship M, from O, North-Eaft 48. oo deg.

2. For, the distance of the Ships-M and O.


As the Sine of MOP, 118. 80 d. (61. 20)

Is to the Distance failed from P to M. 236. 60 d.


So is the Sine of OP M, 36. 00 deg.




To the Distance OM 158.70 M.


Fig. North and South of each other, and are diftant 37.75 HERE are Two Iflands, as K and M, which lie directly


Miles; And there is a third Ifland at L, which is diftant from that at M, 59. 64 Miles; and from that at K, 80. 92 Miles: Now I would know, How the Ifland at L, bears from the other two Iflands Mand K.

The Geometrical Conftruction.

Having drawn NS for the Meridian, and W E for a Parallel of Latitude, croffing each other in the point K, which let be the place of one of the Ilands; Then

The Ifland M, lying directly North therefrom, at the Distance of 37.75 M. Set 37.75 from K to M. And because the Ifland at L is diftant from the Island at K, 8c. 92 M, with that diftance, fet one foot of the Compaffes in K, and with the other defcribe the Arch a b.Also, the diftance of the Ifland at L,

being diftant from M, 59. 64 Mil. with that diftance fet one Fig. foot in M, and with the other defcribe the arched, croffing LXVI.. the former archa b, in L: Laftly, Joyn L M, and LK, fo thall you have conftituted an Oblique-Angled Triangle K L M; in which all the three Sides are Given And the Angles at K and M are Required:

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To the Part of the Greater Side L O, 26. 34. This L O, 26. 34, fubftracted from K L 80. 92 leaves 54. 58 for the other part of the longeft fide KO; and the half thereof, 27. 29 is the part K P or PO. And now, the Oblique-Angled Triangle KLM, is Reduced into two Right-angled Triangles M PK, and MP L; in both which the Hypotenufes and Bafes are given,. by which the other Angles may be found, By CASE III. of R. A. P. T. According to which, you fhall find the Angle MKP. to be 43.71 de. and the Angle MLK 25. 94 deg. L. Bears from K North-Eaft 43. 71 deg.. And M from NorthEaft 69.65 deg


So that

There are two Head-Lands at B and C, which are diftant Fig. 356.00 Miles; And there is a Port at A, from whence two Ships LXVII.: fail to those two Head-Lands, the Ship from A to C, fails South-Eaft 34. 90 deg. and that from A to B, South-Weft 72. 10 de. And both the Ships together have failed 424. 12 Miles.. I demand the Distance that each Ship failed; Or, how far are either of the Head-Lands from the Port at A?

The Geometrical Conftruction.

NE Ship failing South-Eaft 34. 90 deg. and the other South-Weft 72. 10 deg. the diftance of thofe Rumbs is 107. co de. Wherefore,,


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