the Course For the Course As the Distance run A B 108 miles comp arith 7,966576 Is to Radius   So is the Difference of Latitude A D 90 miles 1,954242 To the Course or Rumb Sine comp B 33 d. 33 m. 9,920818 For the Departure As Radius  To the Distance run A B 108 miles 2,0334237 So is Sine the Course A 33 deg 33 min. 9,9424616 To D B the Diff of Longi here found to be 60 miles 1,7758853 You may ask the reason why the Course is not 33 deg 45 min. as it was first given to be I answer because of the part of an unite that the Difference of Latitude was set down more than it should be for if you observe I said it was almost 90 miles when I found it before Now the error is not worth minding in Sailing for the Difference that the Fraction causeth is but 12. min. of a degree of a point of the Compass which is no more than the 56 th part of a point If you mind I have worked the Course thus found and it doth not alter the Difference of Longitude or Departure which should be 60 miles if I had worked to the least Fraction for the Logarithm here is nearest to the Logarithm of 60 in the Tables Indeed I worked on purpose thus because those that I have taught when they could not find it come out right as the other was could not tell where the fault lay now this will direct them to know that some part of a unite missing in the finding of the Sides may make some minutes Difference in the Course But now here in the following Examples we will work to a Fraction namely to the 10 th or 100 th part of a unite in the sides to shew how it is done though in Sailing if you work to half an unite it is near enough or for any ordinary uses and that the Tables give and most commonly to less Departure from the Meridian and Distance run given to find the Course and Difference of Latitude QUESTION V. Admit the Departure from the Meridian were the same that the Difference of Latitude was in the first Question 60 miles the Distance run also 108 miles I demand the Course and Difference of Latitude let the Latitude you set from be 50 deg HEre the Course from the Meridian will be the Complement of the Course there and the Difference of Latitude will be the Difference of Longitude or Departure from the Meridian there But instead of taking the Logarithm answering to the number here take the Logarithm answering to ten times the number and so you shall always have your sides come out to the tenth part of a unite I say it will come out in tenths for you must divide by 10 to bring it into miles after it is done Now instead of multiplying your given sides by 10 do but set a Cipher more to the number and it is done as now for 60 set down 600 which is ten times 60 for 108 set down 1080 which is ten times 108 and thus for any number that will be comprehended in the Radius of the Tables Proportion for the Course As Distance run given B A 1080 tenths comp arith 6,9665762 Is to Radius  So is the Departure from the Meridian 600 tenths D B 2,7781512 To the Sine of the Course A 33 deg 45 min. 9,7447274 For the Difference of Latitude As Radius  To the Distance run 1080 tenths B A 3,0334327 So is the Sine comp of A 33 deg 45 m. 9,9198464 To the Diff. of Latitude in tenths D A 898 tenths 2,9533708 Divide these tens by 10 and it gives the miles contained in the side D A the Difference of Latitude which is 89 miles 8 tenths of a mile whereas before it was 90 miles whis is 2 tenths of a mile more Now when you have the sides thus found in tenths or Centisms work for the Angles and you will find them to come out roundly alike But if you desire more exactness work then in Centisms provided that the numbers in Centisms do not out-run the Tables A Centism is the hundredth part of an unite or of any thing and if I would put the number 60 into Centisms I will set two Ciphers behind it and it is the same as before I multiplied 60 by 100 for it is 6000 the like is to be understood of any other number I would work to the nearest Fraction for the Angles here but I conceive it to be no way beneficial and therefore I 'll refer it to the work of the Sphere where it is of â—uch more use this finds the Angles to a minute which is near enough It is like that some will be so curious that this way of finding things to the tenth or hundredth part of unit will not suffice them but they would have the real number answering to the Logarithm to the 1000 th part of a unite because I find it of no real consideration or use in Navigation I omit that here and desire such to look in Book 1. Page 11. of my Fathers Trigonometry and there it is plainly shewed By the Plain Scale This Question differs nothing from the other in its operation only as you work in the other from a North and South line here you work from an East and West line The work is this First draw an East and West line and from the west end of it set off your Departure from the Meridian which is from D to B 60 miles from D raise the Perpendicular D A which is a North and South line and draw it at length then take the Distance run 108 miles and fix one foot of your Compasses in B with the other cut the Meridian line D A which it doth in A. Draw the side A B then is B the place your Ship is at after her Sailing A is the place of her setting out A D is the Difference of Latitude and N O s is an Arch of 60 deg from the Angle D A B as a Center the measure of N O is the Course and is performed as before in the last Example and also of the Difference of Latitude as in other Examples The Difference of Latitude is D B 90 miles The Course is North 33 deg 45 min. Easterly D A B or N O. The Plain Scale of equal parts will not resolve to the least Fraction as the Tables will now to a tenth part of a mile I confess a good Diagonal Scale will produce things to a small Fraction If you would know what Latitude you are in divide 90 by 60 and the Quotient will be degrees namely 1 deg the Remainder minutes namely 30 min. Now because your Latitude was Northerly and you have gone to the Northwards you have increased it 1 deg 30 min. so that you must add the Difference of Latitude to 50 deg and it makes 51 deg 30 min. if you have gone to the Southwards you must
of the Compass it is to be set off in as also whether it be nearer the East or West than the North or South for if it be nearer the East or West than the North or South you must do the same from a Parallel as here you did from a Meridian for the side of a Square is but the measure of four points which is but half the points between a Meridian and a Parallel This way of working may seem hard and tedious at first but you will soon find that it is free from mistakes and both exact and easie if you practise it Place this between Page 112 and 113. Of OBLIQUE TRIANGLES Two Sides with an Angle opposite to one of them given to find the other Angles and Side QUESTION I. Two Ships set sail from the Rock of Lisbon one sailed W S W the other sailed N W b W 38 leagues and at the end of their sailing they were 58 leagues asunder I demand the Southermost Ships Distance run and how the Ships bear one from the other FIrst draw a North and South line white and then from that set off the Northermost Ships Course make the Rock or Place setting out the Place in the North and South line that you draw that Course from which is C upon this Course set off 38 leagues because the Question saith you sailed 38 leagues upon it and extend the side C A to the Arch of 60 deg at t and set off five points from t to S and draw S C a white line which is W S W Course for it is five points between the W S W and the N W b W this done take 58 leagues from the Scale of equal parts and fix one foot of your Compasses in A and where the other intersects the Course B C which it doth in B there is the other Ship then to measure the Angle of the Ships bearing one from the other it is B and B C is an E N E line extend B A to the Arch of 60 deg whose Center is at B and see how many degrees or points it is more Northerly than an E N E line and so the other Ship namely the Ship at A bears from the Ship at B then take the length of B C and apply it to your Scale and see how many leagues or miles it is I have wrought this in leagues but I will work the rest in miles because it is more exact I find that the Ship at A bears from B N E b N 45 min. Easterly the Distance run of the Southermost Ship is 70 leagues By the Tables The proportion of this and all others of this kind is the same that holds in right angled Triangles namely that the Sine of every Angle is proportional to its opposite side or every side is proportional to the Sine of its opposite Angle Here we have given the side C A 38 Leagues the side A B 58 Leagues and the Angle at C which is 5 points or 56 deg 15 min. Say then for the Angle at B. As A B 58 leagues 580 tem comp arith 7,2365719 Is to A C B 56 deg 15 min. Sine  9,9198464 So is A C 38 leagues 380 tem  2,5797836 To A B C Sine 33 deg 0 min.  9,7362019 Which subtract from 6 points or 67 d. 30 m.  33 00 Remainder is the Course from North 34 d. 30 m. Which the Ship A bears from the Ship B which is N E b N 45 min. Easterly For the Distance run of the Ship at B the Side B C. As Sine C 56 deg 15 min. comp arith 0,080153 Is to A B 58 leagues  2,763428 So is Sine B A C 90 deg 45 min. Take the Sine of the acute Angle B A t 89 d. 15 m. 9,999961 To B C 68 8 10  2,843542 The reason why you take the acute Angle is because the Tables go no further than 90 deg neither indeed is any Sine beyond 90 deg but as my Father saith in his Trigonometry p. 2 the Sine of an Arch less than a Quadrant is also the right Sine of an Arch as much greater than a Quadrant So then the right Sine of the Arch 90 deg 45 min. which is 45 min. greater than a Quadrant must be the right Sine of the Arch 89 deg 15 min. which is as much less namely 45 min. the Geometrical Demonstration of it is there laid down You may ask how I came to know the Angle B A C which was thus I found one of the other two Angles namely B and the other I had given me I added them both together and that Sum I subtracted from 180 deg the Remainder must then be the Angle at A because 180 deg is the Sum of the three Angles of any right lined Triangle And if I subtract two of them from three of them there will remain one of them which was 90 deg 45 min. The three Angles of a Triangle given with one of the Sides to find the other two sides QUESTION II. Admit I set from a Head-land lying in the Latitude of 50 deg 00 min. North Latitude and sail W S W 38 Leagues and then meet with a Ship that came from a Place which lies S S W from the Head-land Now this Ship hath Sailed N W. I demand the Distance of that Place from the other and also the Distance the Ship hath sailed that came from the Southermost Place The distance between the two places is A B 58 miles The distance that the Ship sailed is A O 44 4 10 miles If you have occasion to find the Latitude the place at A lies in let fall the Perpendicular A N upon the South line that comes from B and it cuts it in N and leaves the Difference of Latitude B N see how many miles it is and subtract it from 50 deg and you have your desire measure A N and you have the Westing that A lies from the Head-land B. By the Tables Here the Angle at B is an Angle of 4 points or 45 d. 00 m. The Angle at A is an Angle of 6 points or 67 d. 30 m. The Angle at O is an Angle of 6 points or 67 d. 30 m. The side B O is 58 miles   First for the Side A B. The general Rule saith That the Sine of every Angle is proportional to its opposite side Then from this I conclude that B A should be equal to B O because the Angles opposite to them are equal and so you will find them For as Sine A 67 deg 30 min. comp arith 0,034384 Is to O B 58 miles  1,763428 So is Sine O 67 deg 30 min.  9,965615 To A B 58 miles the two places distance 1,763428 You see it is so exactly for these three numbers added together and Radius cast away produceth the same Logarithm that 58 taken out of the Book did and this Question I do on
which is A C and cuts the the Westermost Head-land Lastly draw out a South line from A which is A D and cuts the Island Lastly A s and R s are the Courses that lead from the two Stations to the Island and therefore s is the Island for there they cut Draw a line from every one of them as from s to O and from O to I from I to s and take the distance of them asunder then for their bearing one from another draw a North and South line from each place and by an Arch of 60 deg measure it as hath been shewed By the Tables First I consider that I have the Distance between the two Stations R A which is a side of the triangles A R I A R O and A R s and in every of these Triangles the Question gives me all the Angles so that I can find my Distance from my Station to the Head-lands after the form of this Example which I shall do from each Station to the Eastermost For A I. As Sine R I A 38 deg 45 min. comp arith 0,20347 To A R 8 miles  1,90309 So is Sine A R I 28 deg 45 min.  9,68213 To A I 6 miles 2 10  1,78869 For R I. As R I A Sine 38 deg 45 min. comp arith 0,20347 To A R 8 miles  1,90309 So is R A I Sine its comp to 180 d. 67 d. 30 m. 9,96561 To R I 11 miles 8 10  2,07217 Here I have found the Distance from each Station to the Eastermost Head-land next find the distance from the Westermost Head-land to the first Station A O. For A O. As Sine R O A 6 points or 67 d. 30 m. co ar 0,034384 Is to R A 8 miles  1,903090 So is O R A 67 deg 30 min. Sine  9,965615 To A O 8 miles  1,903090 Thus you might find R O R s and A s for you have the same things given namely R A and the Angles Here we have found the sides A I and A O and because A I is upon a N N E Course and A O upon a N W Course the Angles contained between O A I must be 6 points or 67 deg 30 min. so that we have two sides and their contained Angle to find the Angles of the bearing of these Head-lands A I O or A O I and the Distance of them asunder O I this is wrought as the third Question is Thus as this is found so you may find the bearing of the rest of the places one from the other if there were twenty of them for you have as much given in the other Triangles but I leave it to your own practice also to make such Experiments or to frame such Questions Two Sides and a contained Angle being given to find the third Side and the Course that each Side runs upon provided that no Course be named only the half of the Compass that you sail in or the quarter which is most commonly and the Difference of Latitude that is made between the extent of the two Sides QUESTION VI. Two Ships make an Angle of 50 deg the one sails between the South and the West 40 leagues the other sails between the South and the East 50 leagues and at the end of their sailing the Eastermost Ship is more Southerly than the Westermost so that they differ their Latitude 15 leagues I demand each Ships Course and their Distance asunder THe place the Ship set from I make to be B take off 60 deg from your Scale and describe an Arch making B your Center as the Arch T R then I take 50 deg from my Scale and set off somewhere upon that Arch namely from T to R and from B upon the Course that leads to T I set off 40 leagues my Westermost Ships Distance and so conclude my Westermost Ship is at C then from B upon the Course that leads to R I set off my my Eastermost Ships Distance run which is 50 leagues from B to A. Lastly draw A C which is the Distance of the Ships at the end of their sailing Then because the Question saith the Eastermost Ship was 15 leagues more Southerly in Latitude than the Westermost I will take 15 leagues from the Scale of equal parts and fix one foot of my Compasses in A and with the other describe the Arch I S N and draw the line from C by the upper edge of the Circle at length which must be an East and West line because A is just 15 leagues to the Southwards of the nearest part of it then let fall a Perpendicular from B upon C N and extend it as far as the Arch of 60 deg which you first described T R and it cuts in O this must be a South line from B namely B S then measure upon the Arch of 60 deg from O to R for the Eastermost Ships Course and from O to T for the Westermost Ships Course as you do in any other Question I find them here as followeth The Eastermost Ship sailed S b E 4. d. 40 m. Easterly The Westermost Ship sailed S W 11 deg Southerly The Distance asunder is C A 39 leagues By Proportion Here are the two sides B A 50 leagues and B C 40 with the contained Angle C B A given to find the Angles A C B and B A C therefore for the Angles As B A and B C 90 leagues comp arith 7,045757 Is to their Difference 10 leagues  2,000000 So is Tan. half the Angles B A C and B C A 65 d. 0 m. 10,331327 To Tang. their Difference 13 deg 24 min. 9,377084 Which added to the half sum 65 deg is 78 deg 24 min. the Angle A C B subtracted from 65 deg it makes 51 deg 36 min. the Angle C A B. Then for the Ships Distance asunder C A. As Sine A C B 78 deg 24 min. comp arith 0,008962 Is to B A 50 leagues  2,698970 So is Sine C B A 50 deg 0 min.  9,884254 To C A 39 deg 1 10 leagues  2,592186 Here we have found the required Angles and the required side but we know not what point of the Compass either of the sides that bounds the Triangles runs upon but this which we have found makes way to it for now we know the side A C to be 39 1 10 leagues and if you let fall a Perpendicular from A upon the East and West line C N it cuts it in S then A S is known to be the Semidiameter of the Arch I N S u 15 leagues so that in the Triangle A S C right angled at S we have A S 15 leagues given and A C 39 1 10 leagues to find A C S or C A S we will find A C S. As A C 39 1 10 leagues 7,407823 Is to Radius or A S C 90 deg  So is A S 15 leagues 2,176091 To Sine A C S 22 deg
the Course A 56 deg 15 min. 9,824897 To the Difference of Latitude in parts 1968 parts 3,293977 This done I look in the Tables against the Latitude that is given me 23 deg 9 min. and the Meridional part answering to it is 1428 To which I add the parts here found 1968 because the Latitude is increased  And there comes forth 3396 I look for that in the Tables and I find answering to it the Latitude of 49 deg 9 min. which is the Latitude of the unknown place B. For the Distance between these Places I subtract the Latitude of A 23 deg 9 min. from the Latitude of B now found 49 deg 9 min. and the Remainder is the Difference of Latitude between them 26 deg 0 min. which converted into miles is 1560 miles then say As Sine comp 56 deg 15 min. A comp arith 0,255260 Is to A B 1560 miles 3,193124 So is Radius  To the Distance run A C 2808 miles 3,448384 I had thoughts to have shewed the Difference between Plano and Mercator in an Example but my Father hath done it in page 170 of his Book If you are desirous to know Great Circle Sailing you may learn it from his Book for he hath treated of it largely and indeed it is the most necessary way of Sailing that is for a man cuts his way much shorter in many runs than the usual way that men go doth If you desire to know in any Latitude the number of miles contained in any number of degrees of Longitude convert your degrees and minutes of Longitude into minutes and say As Radius To Sine comp that Latitude So is the Difference of Longitude in minutes To the Difference of Longitude in miles If you desire to convert miles of Departure in any Parallel into degrees say As Sine comp the Latitude or the Parallels dist from the Pole Is to Radius So is the Departure in miles To the Difference of Longitude in minutes Divide that by 60 and the Quotient with the Remainder is the degrees and minutes contained in that Departure Of the Longitude and Latitude of Places BEfore I come to shew how to keep a Reckoning by Mercator's way I will set down a Catalogue of the Latitude and Longitude of some Places in degrees and minutes the Longitude taken from Floures and Corves Places Names Latitude Longitude D. M. D. M. The Tassel North 53 00 35 10 The Maze North 52 00 34 29 The Willing 51 30 33 53 Isle of Wight 50 24 29 16 Dover 51 05 32 04 Portland 50 24 28 16 The Start 50 07 27 07 The Lizard 50 00 25 20 Silly 50 04 24 00 Vshin 48 30 25 30 Cape Finistre 43 08 20 30 The Rock of Lisbon 38 52 20 47 Cape St. Vincent 37 00 21 29 Gilbraltar 36 01 24 39 Cape Blanco 20 32 13 09 Cape de Verde 14 31 13 10 Serrat Lion 08 00 17 53 South side of St. Ann 06 42 17 32 Cape de Palmose 04 00 24 30 Cape Terras Pantas 04 06 30 30 Cape Formosa 04 04 39 15 River Comrans 03 19 42 32 Nethermost of the Isle Farmandosa 03 26 41 56 Middle of the Isle of Thoma 00 00 40 00 Cape de Lopez 10 00 42 01 Cape Negro South 16 00 4 521 Cape Bonesperanza South 34 25 52 30 North end of Martinass 23 02 17 16 Isle of Picos 23 01 17 16 Isle Terrestinam 37 15 23 31 Y Digon Aluarum 38 54 26 45 Ascension 08 21 20 16 St. Ellna 16 00 28 21 Cape Agulhas 35 00 54 00 Pomte Primere 32 25 63 06 Cape Corentas 24 00 69 50 Mosambique 15 10 75 32 Pemba 04 41 77 02 Cape Derfue 10 20 87 40 Cape Guardafa 12 01 87 41 Cape de Raffas 22 10 97 29 Serrat 21 01 108 35 Goa 15 40 109 22 Cape Cormoram 07 53 112 20 Point Gada 06 02 115 00 Northwest Point of Sumatra 05 30 129 02 The Flat Point 05 51 137 26 Crequeta in Sinda 06 02 137 21 East end of St. Brandon 17 15 102 05 East end of Diego Roderigos 19 24 99 22 Southeast Harbor of Maurice 20 11 92 54 Southeast end of Mastorbos 20 16 90 22 Cape de Roman 25 02 83 24 Isle St. Paulo 38 30 108 40 Isle St. Maria Dogusto 19 02 13 00 Assemaon 20 01 02 42 East end of the Foul Ground 18 21 00 00 Isle de Lobas at Rio de Plato 35 25 343 02 Cape St. Thoma 22 21 358 01 Bay Todas Sanctos 13 15 357 41 Cape St. Augustin 08 41 001 12 Isle Fernando 03 45 005 14 Vizia 03 30 003 01 Venedo St. Paulo 01 52 006 25 The West-Indies Barbadoes North 13 12 327 42 Ditiatha North 16 18 326 32 St. Christophers 17 20 324 52 Tobago 11 16 327 41 East side of Trinidado 11 37 326 41 Mattinino 14 48 325 30 St. Vincent 13 10 326 00 New point of Portorica 19 02 319 22 Altanalla near Hispaniola 17 26 313 52 Cape Tibarune in Hispaniola 18 14 310 22 West end of Margareta 11 11 321 51 Cartagena 10 49 310 41 Island Guanatho 16 51 297 11 West end of Cuba 22 00 298 44 Cape Florida 25 20 302 35 Bermuda Island the body of it 32 20 325 15 Island of Sable 43 40 334 02 Cape Race at New-found-land 46 32 339 32 Cape Sable 43 41 327 21 Cape Cod 41 20 322 01 Cape Hattarass 35 40 314 11 Cape Henry 37 00 344 00 Cape Charles 37 20 313 55 West side of CORVES 40 00 000 00 East side of FLOVRES 37 20 000 00 The Rode of Fial 38 50 002 17 West end of Fial 38 41 002 48 West end of St. George 39 00 003 00 Gratiosa 39 15 003 20 West end of Tercera North 39 00 03 40 East end of Michael North 38 05 06 10 East end of Maria 37 00 06 01 East point of Porto Santo 33 05 15 26 West end of Madera 32 20 13 48 East side of Palma 28 45 12 48 North point of Gomera 28 10 13 20 North side of Ferro 27 40 12 30 north-Northeast side of Gammest 28 36 14 31 north-Northeast point of grand Canaria 28 20 15 10 East end of Fortaventure 20 21 17 18 East end of Lausareta 28 53 17 21 East end of Gratiosa 29 00 17 29 East end of St. Antonio 17 21 04 20 East end of St. Vincent 17 12 05 00 East point of St. Lucia 16 54 05 25 East point of Sal 16 54 07 30 East side of Bonavista 16 10 08 00 East side of the Isle of May 15 00 07 30 East side of St. Jago 15 10 07 02 Fogo 14 36 06 00 Bramma Island 14 24 05 38 The Longitudes and Latitudess of these places here inserted are some of them my own Experiences and the rest Accounts of able men Concerning the foregoing Tables HE that is bound to any place the first thing that he ought to consider is the Latitude of each place and the
6 is equal to the Sine of S ♈ when you have measured it convert it into time and set it down I find it to be 1 hour 44 min. The Meridian Altitude of the Sun being given and the Latitude to find the Suns Declination as also all the other things before found Latitude 51 deg Northerly Meridian Altitude 52 deg 30 min. I demand the Suns Declination or any other thing that I found before FOR the doing of this project your Sphere as before was shewed that done I consider the Meridian Altitude of the Sun is given to be 52 deg 30 min. I will take 52 deg 30 min. from the Scale of Chords and fix one foot of my Compasses in the South point of the Horizon at A and extend the other foot upwards upon the Meridian which falls at B then I conclude the Sun is upon the Meridian in the point B Take the Distance between B and the Aequinoctial which is B C and apply it to the Scale of Chords and so much as it is so much is the Declination of the Sun and you see it is Northerly for the Sun is to the Northwards of the Aequinoctial If you have a desire to find any thing else in this Sphere draw the Parallel of the Suns Declination thus found and work with it as before in other Examples also draw the Ecliptick and Tropick Many other things might be done but for Brevities sake I omit them judging this to be as much as we have occasion for in Navigation which is the thing I aim at If I have committed a fault in being too tedious in these things pardon me and remember it was my Beginnings riper years which may bring more ability it is probable will bring forth something more new and admirable Be charitable Terms of Art used in Navigation LAtitude as I have cited before is the Distance between the Aequinoctial and the Zenith what is meant by Longitude and Difference of Latitude I have shewed before also the difference between Longitude and Departure from the Meridian in this Book before Now in every Distance that is sailed upon any Course that makes an Angle from the Meridian either of North or South or the Parallels of East and West that is to say any Distance that is sailed upon a point that is between those Courses we have these things considerable namely The Distance run which must be the longest side the Difference of Latitude and Difference of Longitude or Departure from the Meridian and the Course the Course is the Point of the Compass or Rumb you steer on A Dead Reckoning is when a man cannot observe but is forced to judge of the Course and Distance to find the Difference of Latitude and Departure A Reckoning is an Account of the Difference of Latitude and Departure from the Meridian that is made in the time a Ship hath been out Read Tan. for Tangent Sine com for Sine Complement Tan. com for Tangent Complement comp arith for Complement Arithmetical PROPOSITIONS of SAILING BY THE PLAIN SCALE Difference of Latitude and Course given to find the Distance run and the Departure from the Meridian QUESTION I. Admit I set from the Lizard lying in the Latitude of 50 deg and sail S W b W till I alter my Latitude 1 degree or 60 miles I demand my Distance run and Departure from the Meridian IN this Question we have the Course 5 points from the Meridian towards the West which is 5 times 11 deg 15 min. or 56 deg 15 min. for 5 times 15 min. is 75 min. or 1 deg 15 min. and 5 times 11 deg is 55 deg to which add the 1 deg and 15 min. and the whole is 56 deg 15 min. To protract this Question or any other of this nature I wish you to observe the end of your Book that lies from you to be the North part of your Book then that to the right hand will be the East part that to the left hand the West and that right to you the South part The Reason why I wish you to observe this Method is that so in any Travis you may set every Angle the right way which must be observed then and is good to be always used But besides this Geometrical way of resolving Triangles there is an exacter way by the Tables of Sines Tangents and Logarithms which is done thus A General Rule Note that in all right lined Triangles the Sine of every Angle is proportional to its opposite side or every side is proportional to the Sine of its opposite Angle The greatest side will have the greatest Angle opposite to it the last side will have the least Angle opposite to it As here in this Question A B is the longest side and opposite to it is the right Angle B D A which is the greatest Angle A D is the next and hath A B D to it D B is the least side and hath D A B for its opposite Angle which is the least Angle Now as the Sine of A is to B D so is the Sine of D to A B or the Sine of B to A D and the contrary To find the Distance run As Sine com B 56 deg 15 min. comp arith 0,25526 To B D 60 min.  1,77815 So is Radius or Sine D 90 deg   To A B 108 miles  2,03341 If Radius be not in the first place you must take the Complement arithmetical of the first number as here I have done but if Radius be in the first place take no Complement arithmetical at all The Complement arithmetical of a number is what every single figure wants of 9 take them in their due places as here the Sine in the Tables answering to A 33 deg 45 min. which is the Sine Complement of B 56 deg 15 min. is 9,74473 whereas I have set it down 0,25526 now if you compare them the number that I have taken is what the figures in the other want of 9 as 9 wants 0 and 7 wants 2 and 4 wants 5 and so forward after the Complement arithmetical of the first number is taken take the others as they be in the Tables and add them together casting away Radius always and look for what comes out either in the Sines Tangents or Logarithms according as the order requires into which the proportion runs You see here I took no notice of Radius which is a casting of him away in like manner if the things added up come to above Radius always cast Radius away and set down what is above Radius as here the last came out to be 11 and I cast away the 10 and set down 1. For the Departure from the Meridian As Sine comp B 56 deg 15 min. comp arith 0,25526 Is to B D 60 miles  1,77815 So is Sine B 56 deg 15 mim  9,91984 To A D almost 90 miles  1,95325 But I have been often asked
purpose to shew the truth of the general Rule The same way other Questions of this nature are wrought For the Distance that the Southermost Ship sailed A O. As A O B Sine 67 deg 30 min. comp arith 0,034384 Is to A B 58 miles  2,763428 So is Sine A B O 45 deg 00 min.  9,849485 To A O 44 4 10 miles  2,647297 If you desire to find the Difference of Latitude between the two places B and A you have the Distance A B and the Course A B n given to find it or the Longitude If you desire to find the Latitude the Ships are in when they meet let fall a Perpendicular from O upon the South line which is B N and you have the distance O B given 58 miles and the Angle O B R 67 deg 30 min. to find it or the Departure between the Ships at their meeting and the Head-land O r. The like is to be understood of any other Question of this nature Two Sides and their contained Angle being given to find the third Side and the other Angles QUESTION III. Two Ships set from one Port and make an Angle of 58 deg one sailed N b E and the other sailed N W 1 deg 45 min. Westerly the Eastermost Ship sailed 70 miles and came to her Port The Westermost Ship sailed 89 miles and came to her Port I demand the Distance of these Ports asunder and how they bear one from the other FIrst draw the North and South line A S. Let A B represent the Port you set from Then from A as a Center describe the Arch of 60 deg F S D From S to the Eastwards set off one point for N b E the Eastermost Ships Course draw the line A D at length and set the first Ships distance 70 miles upon it which is A B then from A draw a N W line 4 deg 45 min. Westerly and set off the Westermost Ships distance run upon that which is 89 miles and it reacheth from A to C draw the line C B from the ends of the other two sides which is the Distance between the Ports for the ends of each Ships run must be the Ports then is A C B the Angle of the bearing of the Eastermost Port from the Westermost or A B C the bearing of the Westermost Port from the Eastermost C A is a S E line 1 deg 45 min. Easterly from C see how many points are contained between the C B and C A by the way that is shewed in the Question before or because B A is N b E line from B you may find the Point of the Compass that B C runs upon from B you need find it but one of these ways and the opposite point must be the other as I have shewed before then take the length of the side C B which is the distance of the Ports asunder I find their distance and bearing to be as followeth The Distance between the two Ports is C B 79 miles The Bearing of them is East 5 deg 35 min. Northerly or West 5 deg 35 min. Southerly By the Tables Here is given the Angle at A 58 deg 00 min. the side A B 70 miles the side A C 89 miles which is two sides and their contained Angle and here is required A C B or A B C and the side B C. First For the Angles For all Questions of this nature this Rule is beneficial and will work your Question Note That in all plain Triangles the Sum of two Sides are in such proportion to their Difference as the Tangent of the half Sum of their opposite Angles is to the Tangent of an Angle which Angle shall be the Difference between the half Sum and the Angles so that if it be added to the half Sum it shall make the greater Angle if it be subtracted it shall make the lesser I have said something of this Rule before which makes me demonstrate it no farther here Proportion As the Sum of A B and A C 159 miles comp arith 6,79860 Is to their Difference 19 miles  2,27875 So is the Tan. of the half Sum of B and C 61 d. om 10,25624 To Tangent of an Angle 12 10 9,33359 This added to the half Sum makes the Angle at B 73 d. 10 m. subtracted it makes the Angle at C 48 deg 50 min. which is 4 points and 3 deg 50 min. then say I if C A be a S E Course 1 deg 45 min. Easterly from C 't is certain that C B must be 4 points and 38 deg 50 min. from it which is East 5 deg 35 min. Northerly the bearing of the Eastermost Port from the Westermost To find the half Sum of the Angles Note that the three Angles are 180 d. therefore if one of them be known subtract that from 180 d. and the Remainder must be the other two Now for their Distance asunder C B. As Sine A C B 48 deg 50 min. comp arith 0,123321 Is to B A 70 mil.  2,845098 So is Sine B A C 58 deg co min.  9,928420 To the Distance of the Ports asunder C B 78 9 10 2,896839 The three Sides of a Triangle given to find the Angles QUESTION IV. Two Ships set from one Place one sails N b E 70 miles the other sails between the North and the West 89 miles and at the end of their sailing they are distant asunder 78 miles 9 10 of a mile I demand what Course the Westermost Ship sailed and how the two Ships bear one from the other FIrst draw the North and South line A e make A the place of the Ships setting out set off a N b E Course upon the Arch of 60 deg I e which is e D because the Eastermost Ship sails N b E likewise set off her Distance which she sailed upon that Course which is A B 70 miles then take the other Ships Distance which is 89 miles and fixing one foot of your Compasses in A with the other describe the Arch at C Lastly take the Distance of the two Ships asunder 78 miles 9 10 and fixing one foot of your Compasses in B cross that Arch and from the place of intersection draw the sides of the Triangle C A and C B then measure the Angles from an Arch of 60 deg as hath been shewed before and to know them consider that A B is a N b E line so that A F will be upon a Course so many points or degrees and minutes from a N b E as there are points or degrees and minutes in the Arch D F. Secondly for the bearing of the Ships which is the Angle at B consider that B A is a S b W line from B and then the degrees and minutes or points contained in the Arch O R is so many degrees and minutes or points from the S b W. I have resolved the Demands which are as followeth The Course that the