of their intersection draw your line to the end of your given line and it shall be a Perpendicular you see it is the same the other was PROP. IV. To let fall a Perpendicular from a Point to a given Line LEt the given line be D A the point from whence the Perpendicular is to be let fall be at C. From the point C draw a white line to the given line by guess as C A divide it into two equal parts which is done at B then continuing ½ the line C A which is A B or C B in your Compasses and your Compasses fixed one foot at B describe the Arch C D and where it cuts the given line there will your Perpendicular fall from the given point for C D is Perpendicular to the given line D A. PROP. V. To draw a Line parallel to a Line given LEt the given line be A S It is required to draw a line so that the two lines may run at both ends one by the other and never meet which is parallel one to the other Open your Compasses to that extent as you would have the two lines asunder and go towards one end of the given line as at S and describe the Arch u and with the same distance come towards the other end as at A and describe another Arch which is N and by the top of these two Arches draw the line R O which is parallel to A S. PROP. VI. To draw a Line parallel to a given Line from any Point assigned LEt the given line be S L the point assigned be A take the distance from S to A and carry it towards the other end of the given line as at L describe the Arch n then take the distance from L to S and fixing one foot of your Compasses in the given point A cross the Arch n with the arch o and by the place of their intersection and the point assigned draw a line which shall be parallel to S L. PROP. VII To make a Square of a Line given LEt the Line given be A equal to which draw the side of the Square B E and from one end of it raise a Perpendicular and by it set off the length of A as here from E the Perpendicular was raised and the length of A set upon it which is the side of the Square E D continue the same distance in your Compasses and go to D and describe the Arch 8 carry the same distance to B and cross the Arch 8 with f and from the intersection of those two Arches draw the sides C D and C B which makes the Square BEDC this is a true square PROP. VIII To make a Square whose Length and Bredth is given THese sorts of Squares are called Geometrical Squares when but two sides are equal namely the two longest sides or the two shortest sides The Angles are all equal namely right Angles Suppose the Length of the Square be A the bredth B I desire to make it first draw a line equal to A for the length of it as S V then from any end of that Line raise a Perpendicular as here from V and set off the line B for the bredth of it upon it which falls in the Perpendicular line at L then take the length of A and describe the Arch n fixing your Compasses one foot in L then take B the bredth and cross that Arch by another fixing your Compasses one foot in S and draw L K and K S from the place of their intersection as you did in the other Thus the two opposite sides in this Square are equal and the Angles in both all equal for they are right Angles PROP. IX To make a Diamond Figure of a Line and an Angle given A Diamond Figure is a Figure of four equal sides but the Angles are two of them acute and two of them obtuse the acute Angles are equal and the obtuse Angles are equal one to another Let A B be the given line C B the measure of the given Angle A being the angular point first take the line AB and draw a line of its length for one side of the figure namely 8 0 then take the Semidiameter of the Arch C B which is C A and fixing one foot of your Compasses in 8 describe the Arch S 0 and take the Arch B C and set it off from O to S then draw the line S 8 equal to 8 O this done keep the length of O 8 in your Compasses and from S and O describe the Arches n and t and draw the sides O R and S R as you did in the other figures and thus S R is equal to 8 O or R O is equal to 8 S and the opposite Angles also equal I forbear to shew the reasons of their being equal because it hath been handled by others and indeed it is so plain that with a little consideration you may know it PROP. X. To make a Rhomboiades of two given sides and an Angle included A Rhomboiades is a figure whose opposite sides and opposite Angles are equal as a Geometrical Square is but in this they differ a Rhomboiades hath never a right Angle but two obtuse and two acute whereas the other hath all right Angles it differs from the Diamond figure also for in one the sides are all equal and in this but two equal sides I need not shew the working of it because it differs not from a Diamond figure only in taking the two sides apart to describe the Arches at L I suppose you may conceive how it is made by seeing this which is here made the given sides and Angle is s O N I have not set down the Arch to measure the Angle at O. I suppose from what hath been said you will conceive how that is PROP. XI To find the Center of a Circle Draw a line from side to side of the Circle at a venture as A C and divide that line into two equal parts by a Perpendicular as was shewed before that Perpendicular line draw through the Circle from side to side as is u S and it shall be the Diameter of the Circle the half of which is the Semidiameter or very Center ⊙ It is possible to find the Center of a Triangle after the same manner Suppose the Triangle whose Center you would find were A C n divide any side into two equal parts by a Perpendicular and it will go through the Center of the Triangle as the side A C is divided into two equal parts by the Perpendicular B S then I say B S goeth through the Center of this Triangle but to find whereabouts in this Perpendicular the Center of the Triangle is I know by no other means but by removing your Compasses in this line from place to place till you find it which is here found to be at u. But this is but a botchingly way and with a little more labour you may find it at once therefore mind
to good reason that being the Angles in this Triangle are the same with the Angles in the Question therefore as the sum of these sides are in proportion to the Sum of the sides there so is the sum of the side 30 here to the sum of the side that is correspondent to it there or runs upon the same Course To find the Ships Distance that sailed S W correspondent to C N 30 leagues As sum of the sides of the Triangle C N B 95 le co ar 7,021363 Is to the sum of the sides in the Quest 148 leagues 3,170261 So is the side C N 30 leagues 2,477121 To the Ships Distance that sailed S W 46 leagues 6 19 2,668745 This side being thus found we to abbreviate the work will lay it down in the Triangle before whose Angles are equal to to the Angles here set 46 6 10 leagues for the length of the side the Westermost Ships run which is N R. Draw R s parallel to C B and extend N B till it intersect R s which it doth in s This is as much as protracting of it anew for the Angles at R at s and at N are the same that they were given to be in the Question and the side R N 47 leagues as it was found You may measure the sides unknown by the Plain Scale and set them down if you will or you may work by the Rule of Three using this Proportion As N C 30 leagues is to N R 46 6 10 leag So is N B 41 6 10 to N S. Multiply and divide and it maketh 64 6 10 leagues 14 57 of a Tenth By the Tables for N s As N C 30 leagues comp arith 7,52287 To N R 46 6 10 leagues  2,66839 So is N B 41 6 10  2,61909 To N s 64 6 10 leagues  2,81035 We might have done this by the Tables putting the Angles and Sides in proportion as we have done all along in the other Triangles But I suppose your own reason will give you that this hath the same given in it that the Angle C N B had and is wrought so The Fractions here and the other agree only this is not so true altogether because here we do it but to tenths there in smaller Fractions but this is within a small part of one tenth of a unite For the Side R s As Sine N s R 45 deg 00 min. comp arith 0,15051 Is to R N 46 leagues 6 10  2,66839 So is Sine N 33 deg 45 min.  9,74473 To R s 36 6 10 leagues  2,56363 If you add the sides now found all together it will come to 2 10 of a unite less than the given sides together in the Question is now that ariseth by neglecting the taking of the absolute number answering to the Logarithm that comes out but this is sufficient in cases of this kind If you had set it in paces and wrought to the tenth part you would have been out but 2 10 of a pace therefore use your mind in such cases If you had wrought to Centisms you would have been nearer but this I count near enough for sailing and I am sure is more proper to be used than such small Fractions because no long distances can be guessed to a mile or a league You might have wrought your last proportion by the Rule of Three as you did the other Three Sides of a Triangle given to find the Center QUESTION IX There be three Ships bound to one place the Eastermost is distant from the middlemost 40 leagues and bears S E b E the middlemost is distant from the Westermost 50 leagues and bears N E now they every one know as much as I have writ they also know they are of a like distance from this Port. I desire to know what distance the Port is from them and how it bears from each Ship I Have applied this Question to sailing here but before in this Book I have used it in a thing which is very proper for it and indeed it may be of use many times which makes me give it place here I shall say nothing of this Question by the Plain Scale for it is done as before you see but we will do it by the Tables of Tangents and Logarithms Here I have given the side M er 40 leagues the side M W 50 leagues and the Angle at M namely its Complement to 180 deg R M er 78 deg 45 min. to find the Angles M er W and M W er As W M add M er 90 leagues comp arith 7,04575 Is to their Difference 10 leagues  2 So is the Tan. of ½ M er W and M W er 39 d. 22 m. 9,91404 To Tan. of their Difference 5 deg 12 min. 8,95980 Subtracted from 39 deg 22 min. and the Remainder is M W er 34 deg 50 min. added to it makes M er W 44 deg 34 min. For the Side er W. As Sine M er W 44 deg 34 min. comp arith 0,153824 To W M 50 leagues  2,698970 So is Sine R M er 78 d. 45 m. its comp log 180 d. 9,991573 To W er 69 9 10 leagues  2,844367 But now you will demand of me how we shall find any thing in the Triangle W F er being there is but one thing known in it namely the Distance of the Eastermost Ship from the Westermost er W I 'll tell you Mr. Euclid proves this which makes me forbear it Mr. Euclid Book 3. Prop. 20. saith That in a Circle an Angle at the Center is double to an Angle at the Circumference provided that both the Angles have to their Base one and the same part of the Circumference The Center is F here the Triangle W F er and W M er have W er to both their Bases then must B F A be double to I M O or C B A double to D M I that is to the Arch I O D so that if you double D O I 101 deg 15 min. it makes C B A 202 deg 30 min. that subtracted from a Circle 360 deg leaves the Triangle W F er It is evident then that W F er is double to R M I because R M I is what I O D wants of a Semicircle as W F er is what C B A wants of a whole Circle which is double to it Example  deg min. W M er is 101 deg 30 min. doubled it is C B A 202 30 That subtracted from a Circle 360 00 The Remainder is W F er 157 30 And this subtracted from 180 00 The Remainder is F W er and F er W for the Angles of a right angled Triangle are 180 d. 22 30 The half of which is F W er or F er W 11 15 For they must be equal because the sides opposite to them are equal namely F er and F W. And thus we have the three Angles of the Triangle W F er and the side W er to
find F er or F W or F M the distance of any of the Ships from the Port. For the Distance of the Ships from the Port. As Sine W F er 22 deg 30 min. comp arith 0,417160 Is to W er 69 9 10 leagues  2,844477 So is Sine F W er 11 deg 15 min.  9,290235 To any of the Ships Distances from the Port which I find to be 35 6 10 leagues 2,551872 I need not count the Triangle for the points for I have shewed that sufficiently So that you I suppose can tell how to find the Ships bearing from the Port for it differs nothing from the way you count in other Questions QUESTION X. There was two Ships set from one Port one sailed N W b N 6 leagues a watch the other sailed N E 8 leagues a watch these Ships arrived at two several Ports in one instant of time the Ports were 12 leagues asunder I demand the Distance run of each Ship and the Ports bearing one from the other FOr the doing of this Question First draw a N W b N line and set off 6 leagues upon it which is A I then draw a N E line and set 8 leagues off upon it which is A D draw the side I D then consider that as A I is to A D so is the Westermost Ships run to the Eastermost so that the sides A D and A I extended so far that a line of 12 leagues long drawn parallel to D I will just touch the extended sides which must cut them in the places were the Ports are namely A B in B and A I in C measure the sides C A and A B and also the Angles of the Ports bearing A B C or A C B and count them as hath been shewed before I find the side A B the Eastermost Ships run to be 10 leag 7 10. I find A C the Westermost Ships run to be 8 leagues The Westermost port bears from the Eastermost West 4 deg 14 min. Southerly By the Tables First for the Ports bearing one from the other A D I equal to A B C or A I D equal to A C B they are equal because I D and C B are parallel and I A D is an Angle opposite to them both As A D and A I 14 leagues comp arith 6,853871 Is to their Difference 2 leagues  2,301030 So is Tan. ½ A I D and A D I 50 deg 37 min. 30 sec 10,085698 To Tan. their Diff. 9 deg 51 min.  9,240599 Which added to the half Sum is A I D 60 deg 28 min. A C B equal to A D I or A B C the bearing subtracted from the half sum is 40 deg 46 min. For I D. As Sine D 40 deg 46 min. 0,185100 To I A 6 leagues 0,778151 So is Sine I A D 78 deg 45 min. 9,991573 To D I 9 Leagues 0,954824 For C A. As I D 9 leag co ar 8,04575 Is to C B 12 leagues 2,07918 So is I A 6 leagues 1,77815 To A C 8 leagues 1,90309 For A B. As I D 9 leag co ar 8,04575 Is to C B 12 leagues 2,07928 So is A D 8 leagues 1,90309 To A B 10 7 10 leagues 2,02802 The reason that the proportion stands thus is because the Angles are equal in each Triangle and therefore the sides must be proportional Thus the Eastermost Ship sailed 10 7 10 leagues the Westermost 8 leagues and the Eastermost Port bears from the Westermost East 4 deg 14 min. Northerly There is no Distance that a man sails but if he returns the same way that he went out he shall find the same Distance backwards as he did out But now the World is round and if a man goeth out one way and returns another as it is common he will find a great deal of difference in long runs For to demonstrate the reason of this I shall not because my Father and others have done that sufficiently besides it would take up too much room in this little Book These Questions of Sailing may be understood in such cases as this is namely when a man sails the same way home as he doth out But because the Winds will not suffer a man to do so it is necessary to know some better way to sail by Let a man return never so contrary to what he sailed outwards yet at all times to know how near or how far he may be off Now there is no way better than Mercator's way of Sailing it is very excellent and hath been treated of by many and that makes me presume to think that I may treat of it as well as others And first here are the Tables of Meridional Parts by which you are to work A TABLE OF MERIDIONAL PARTS To every Third MINVTE Latitud Mer. Parts D. M. 0 3 3  6 6  9 9  12 12  15 15  18 18  21 21  24 24  27 27  30 30  33 33  36 36  39 39  42 42  45 45  48 48  51 51  54 54  57 57 1 0 60  3 63  6 66  9 69  12 72  15 75  18 78  21 81  24 84  27 87  30 90  33 93  36 96  39 99  42 102  45 105  48 108  51 111  54 114  57 117 2 0 120  3 123  6 126  9 129  12 132  15 135  18 138  21 141  24 144  27 147  30 150  33 153  36 156  39 159  42 162  45 165  48 168  51 171  54 174  57 177 3 0 180  3 183  6 186  9 189  12 192  15 195  18 198  21 201  24 204  27 207  30 210  33 213  36 216  39 219  42 222  45 225  48 228  51 231  54 234  57 237 4 0 240  3 243  6 246  9 249  12 252  15 255  18 258  21 262  24 264  27 267  30 270  33 273  36 276  39 279  42 282  45 285  48 288  51 291  54 294  57 297 5 0 300  3 303  6 306  9 309  12 312  15 315  18 318  21 321  24 324  27 327  30 330  33 333  36 337  39 340  42 343  45 346  48 349  51 353  54 355  57 358 6 0 361  3 364  6 367  9 370  12 373  15 376  18 379  21 382  24 385  27 388  30 391  33 394  36 397  39 400  42 403  45 406  48 409  51 412  54 415  57 418
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