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
a great Circle for by great Circles is the Sphere measured QUESTION I. To find the Meridian Altitude of the Sun IN this or any other Sphere M G E is that part of the Heavens that is visible the other half invisible to us for it is parted from our sight by the Earth and Sea and the furthest part of it which seems to mix as it were with the Heavens we call the Horizon which is the great Circle M E for it is a great Circle though here we are forced to represent it by a streight Line M is the South point of the Horizon Now the Suns Meridian Altitude is his distance between that point and the place he cuts the Meridian that day which is M I fix your Compasses in M and extend the other foot to I and apply it to your Scale of Chords and as many degrees as you find it there so many degrees is the Sun high when he is upon the Meridian that day which is the thing required Note that at O the Sun riseth at P it is 6 of the Clock and P I is equal to ♈ K when it is extended to a great Circle and both the Sine of 90 deg which is extended to a Chord must be the Chord of as many degrees which is 6 hours in time The time between 6 a Clock and 12 which proves that M I is the Meridian Altitude of the Sun and this measuring any Distance from the Meridian is called The first Way of Measuring QUESTION II. To find the Suns Amplitude of Rising and Setting FIx one foot of your Compasses in the place of the Suns Rising which is O and extend the other foot to the Center of the Sphere which is termed the East point of the Horizon and this Distance apply to your Line of Sines if you have any but if you have no line of Sines extend it to the Chord of the same Arch thus Fix your Compasses with that distance one foot in the Meridian so that the other may just sweep a line that goeth through the Center of the Sphere then say I the Arch between that foot of your Compasses that stands in the Circle and the place where this line you sweep cuts the Circle is the Chord of the thing required and will be the same number of degrees upon the line of Chords as O R would be upon the line of Sines and after this manner is the Chord of any distance taken from a line that goeth through the Center of the Sphere found and this is called The second Way of Measuring I find that the Sun riseth 20 deg 53 min. to the Northwards of the East for she hath North Declination and the Latitude is Northerly or sets so much to the Northwards of the West QUESTION III. To find the Suns Azimuth at six of the Clock THe Suns Azimuth at 6 of the Clock is the nearest distance between the Sun at 6 of the Clock and the East and West Azimuth which is P z Now if you mind it P z is taken off from a Circle which is not so great as the Horizon and yet is parallel to it as the line n w is parallel to the line M E and as many degrees as the Sun is from the nearest part of the East and West Azimuth in the little Circle so many he is from it if an Azimuth where drawn in the great Circle for P z is as many degrees in the little Circle as n ♈ is in the great one Now because your Scale is made by the great Circle therefore extend the distance taken from the lesser Circle to the measure of the greater which is done thus Take half the length of the pricked line which is z w and fixing one foot of your Compasses in the Center of the Sphere describe an Arch from some line that goeth through the Center of the Sphere as the Arch M l n then set the distance z P as a Sine upon that Arch for it is a Sine upon that Arch as well as o ♈ was a Sine upon the Meridian I find the Sine of it is the Sine of the Arch l M lay your Scale from the Center of the Sphere by l and draw the line l q then shall q M be the same quantity of degrees upon the great Circle as l M is upon the little one therefore take M q and apply it to the Scale of Chords and it answers your desire And thus is the Suns Azimuth at 6 of the Clock P z found to be 8 deg 36 min. to the Northwards of the East and this is called The third way of Measuring to measure any distance from a line that doth not go through the Center which must represent a small Circle And thus you may find the Suns Azimuth at any time of the day QUESTION IV. To find the Suns height at six of the Clock THe Sun is at P at 6 of the Clock fix one foot of your Compasses in P and extend the other to sweep the Horizon which is the same as though you let fall the Perpendicular P n set it off by The second way of Measuring as was shewed in Quest 2. and apply it to your line of Chords and the reason is because P n represents a part of a great Circle and so is to be understood to be of the nature of those lines that go through the Center of the Sphere for all Azimuths pass through the Zenith and Nadir which are two opposite points I find the Sun is 10 deg 7 min high at 6 of the Clock QUESTION V. To find the Suns height being due East or West VVHere the Suns parallel of Declination cuts the East and West Azimuth is the place the Sun is in when he is due East in the morning for you see he is then over the point of East in the Horizon which is ♈ or the Center therefore take the distance between that place and the Center which is S ♈ and apply it to the line of Sines but if you have no line of Sines extend it to a Chord after the manner of the second Question which I call the The second way of Measuring for it is the distance of a line which goeth through the Center I find the Suns height being due East or West is 17 deg 25 m. Note that the same height that the Sun is being over the East point so high he is being over the West point in the afternoon QUESTION VI. To find the Difference of Ascension THe Difference of Ascension is the portion of time that is between the Suns Rising and six of the Clock If the days be longer than the nights the Sun riseth before 6 but if shorter after 6 but whether it be before or after 6 that he riseth so many hours and minutes as it is from 6 so much is the half day or night longer or shorter than 6 hours from whence it is evident that if the Sun riseth due East he
in the work at all And indeed my desire to be brief makes me omit things that I think may be understood without treating of them I 'll only touch upon an Example in a Sphere which hath South Declination Latitude Northerly 50 deg 00 min. Declination 13 deg 15 min. Southerly Also if you subtract this Amplitude V ♈ from ♈ H 90 deg the remainder is the Suns Azimuth from the South V H or if you add V ♈ to ♈ A 90 deg it is V A the Suns Azimuth of rising from the North forasmuch as V A is as much above 90 deg as V H wants of it The same is to be understood in any Sphere that the Amplitude and Azimuth of the Sun are Complements one of an other to 90 deg in that Quarter When the Suns Declination is Southerly in any Northern Latitude the days are not so long as the nights so that the Sun cannot be up at 6 of the Clock which is at R therefore the Suns Azimuth at 6 of the Clock is of no use Neither is he up when he passeth the East point of the Horizon which is at P so that you cannot take his height but you may measure how far he is under the Horizon at either of those times or you may find his true Azimuth at 6 which is of no value to us that are Seamen because we cannot have a Magnetical Azimuth they are measured as before The Difference of Ascension is here after 6 of the Clock as much as it was before 6 in the last Example And the reason is because the Latitude is the same and the Declination of the Sun is just as far Southerly as before it was Northerly so that the Sun will now be just as long before he riseth after 6 as before he rose sooner it is V R take it off by the third way of Measuring as you did before and set it down I find it to be 1 hour 5 min. 3 15 it serves for the same uses that it did in the other Example For the length of the day subtract the Difference of Ascension V R from 6 hours which is R f and the remainder is f V the length of the Forenoon which doubled is the length of the whole day that subtracted from 24 hours is the length of the night For the hour of the Suns being due East it is P R which is from the place where the Sun cuts the East and West Azimuth to 6 of the Clock and set it off by the third way of Measuring convert your degrees and minutes into time and subtract that time from 6 hours and it gives the due time of the Morning that the Sun cuts the East point of the Compass the reason why you subtract is because the Sun is due East so long before 6 of the Clock Subtract the hour of the Suns being due East from 12 hours and it will give the true time of the Suns being due West For the time of day breaking draw the line 17 ⊙ parallel to the Horizon and 17 deg under it as was shewed before and measure from P to R that is from the place where the Suns Parallel of Declination cuts that Circle of 17 deg under the Horizon to 6 of the Clock convert it into time and subtract it from 6 of the Clock and it leaves the time of Day breaking namely the time in the morning that the Sun is in P you subtract because the day breaks so long before 6 of the Clock If it where so that the Sun were in the Tropick of Capricorn you must take the Distance S q. You see the Sun is past the hour of 6 before it is break of day now in this case you must add whereas before you subtracted For the continuance of Twilight it is measured as before from the place where the Parallel of the Suns Declination cuts the Circle of 17 deg to the place where it cuts the Horizon convert it into time and set it down In this Example it is P V. What I have not mentioned here I have given sufficient instructions about in the other Example before this Latitude 51 deg 30 min. Northerly Declination 11 deg 10 min. Northerly I demand the Suns place and right Ascension BEfore we can find the Suns place in the Ecliptick you are to consider that the Ecliptick is divided into 12 equal parts by the 12 Signs six of these Signs divide that half of the Ecliptick which is to the Northwards of the Aequinoctial and are called Northern Signs and the other 6 divide the half that is to the Southwaads of the Aequinoctial and are called Southern Signs as I have set them down in this Book before with their Names and Months to their Characters and have set the Northern Signs by themselves and the Southern Signs by themselves This plain Superficies can shew but one part of the Globous Body but you must know it is round which makes the Ecliptick to be divided on both sides from ♈ to ♋ is 90 deg from ♋ to ♎ is 90 deg from ♎ to ♑ is 90 deg and from ♑ to ♈ is 90 deg every one of these quarters containing 3 Signs The Ecliptick being thus divided into Signs we may find the Suns place and right Ascension THe Suns place in the Ecliptick is the nearest Distance between the next Aequinoctial point and the Sun in the Ecliptick by Aequinoctial points is meant ♈ and ♎ the two points of intersection that the Aequinoctial and the Ecliptick make Now in this case the Sun must be nearest the Equinoctial point at ♈ because the Month is belonging to a Sign nearer ♈ than ♎ Therefore fix your Compasses in the point of ♈ and extend the other foot to the Sun in the Ecliptick which is I and apply it to the line of Sines or if you have no line of Sines convert it to a Chord by the second way of Measuring and as many degrees as it is so far is the Sun distant from the nearest Aequinoctial point if it exceed 30 deg the Sun must be in that degree of ♉ that is above 30 deg if it exceed 60 deg the Sun must be in that degree of ♊ above 60 but here I find it in less than 30 namely 29 deg 3 min. therefore I conclude the Sun is in 20 deg 3 min. of ♈ But suppose it were time of year that the Sun were returning from the Tropick towards the Aequinoctial and were the same Declination then would the Suns place be the same from entring into ♎ that it now is in ♈ namely in 57 min. of ♠which wants 29 deg 3 min. of entring into ♎ Thus much for the Suns place The Suns right Ascension is measured from the place where the Suns parallel of Declination cuts the Ecliptick to 6 of the Clock which is I 6 set it off by the third way of Measuring to extend it to a Great Circle for the Sine of I
the Difference of Latitude in the whole run 13 leagues 3 10 if you had West Longitude you must have subtracted thus likewise and that which was greatest the remainder belongs to OF A RECKONING FRom what hath been already shewed you understand that if the Latitude of two places be known and the Difference of Longitude between them you may find the bearing of them one from the other their distance asunder or any thing you desire and this is all so plain that if I set from a place and am bound to a place whose Latitude is known and the difference of Longitude also known between them and keep a reckoning of the departure both East and West that I make setting them down in two distinct Columns it is but subtracting one from the other and the remainder will be the East or West Longitude that I have made that of the two which was most also if I know what Latitude I am in and what Latitude the place lies in it is but subtracting the lesser from the greater and the remainder will be the difference of Latitude between your ship and that place provided the Latitudes be both one way Also subtract the Difference of Longitude that you have made provided that it be that way which shortens your Longitude between the place you set from and are bound to from the whole difference of Longitude between the two places and the remainder will be what is still between you and the place you are bound to but if your departure which you have made hath increased the difference of Longitude between the place you set from and are bound to add the remainder after your subtracting the Columns one from the other to the whole difference of Longitude between the places the like for Latitude Example in the Reckoning following Suppose the Island of Ditiatha lies so that it hath been found to have 950 Leagues departure West from the Meridian of the Lizard according to the Courses that be ordinarily sailed by Plano the Latitude of it is 16 deg 16 min. North Latitude the Lizard lies in the Latitude of 50 deg 00 min. Now upon some Occasion that fell in the term of our Voyage namely the 29th of January on Tuesday the Captain demands of me what Distance Ditiatha was from me by Plano how it bears what Leagues of Departure I have to it and the like I look in my Reckoning and I find in the West Column 434 Leagues I look in the East Column and see 60 Leagues I subtract 60 my whole East Column from 434 the whole West Column and the Remainder is 374 Leagues which signifieth the Ship hath departed from the Meridian of the Lizard 374 Leagues West for the West Column was greatest Subtract this West departure 374 Leagues from 950 the whole Departure and I find there is still wanting 576 Leagues This 576 Leagues is the Departure that is from the Meridian that my Ship is in and Ditiatha Subtract also the Latitudes one from the other namely the Latitude your Ship is in and the Latitude of Ditiatha and the Remainder is the Difference of Latitude in degrees and minutes between the Ship and the Place you are bound to which in this Example I find to be 6 degrees 24 minutes Take the Latitude out of the Column of Latitude for that day in the Reckoning and you will find it so if you subtract 16 deg 16 min. from it Thus you have the Difference of Latitude between you that day and your Port and the Departure that the Port hath from the Meridian that you are in that day given you to find the things demanded I will not work it for it hath beeen shewed sufficiently already I have here following set down a Reckoning and afterwards how I work it and lastly the reason that makes me use this way of keeping my whole Reckoning in the last line so that I am not troubled to add it up when I give account of it January the 3d Anno Dom. 1655 we departed from the Lizard lying in the Latitude of 50 d. 00 m. being bound for Ditiatha which lies in the Latitude of 16 d. 16 m. it having West Departure from the Meridian of the Lizard 950 Leagues according to the courses which we then steered at 4 of the clock the Lizard bore from us N N E about 6 leagues distance by estimation EVery day at Sea you set down at noon that 24 hours work Now I advise you to set it down as it is here In the first Column on the left hand I set the day of the Month in the next the day of the Week in the third the Latitude I am in by dead Reckoning or by Observation only I make a distinction between them by the letter E which stands for Estimation and is set down only when I do not observe that so you may know where to begin to correct In the 4 th Column I set down the Easting in the fifth column the Westing The manner that I set my Easting and Westing down is thus Every day there is the Ships course minded and an observation whereby you have the difference of Latitude and course to find the departure or else if you cannot observe you have the distance run guessed at and the course to help your D. M. Week days Latitude Dep. E. in leag Dep. W. in leag deg min. 4 Friday 49 00 000 021 5 Saturday 47 55 000 059 6 Sunday 46 20 000 085 7 Munday 45 00 000 088 8 Tuesday 43 E 54 000 108 9 Wednesday 42 4 000 147 10 Thursday 41 E 19 000 262 11 Friday 41 10 000 264 12 Saturday 40 46 000 274 13 Sunday 40 45 000 178 14 Munday 39 E 49 000 179 15 Tuesday 37 E 24 000 200 16 Wednesday 35 59 000 229 17 Thursday 34 42 000 249 18 Friday 33 15 000 278 19 Saturday 31 15 000 322 20 Sunday 29 5 000 364 21 Munday 28 E 20 000 400 22 Tuesday 27 5 012 400 23 Wednesday 27 30 022 400 24 Thursday 27 22 031 400 25 Friday 26 E 49 049 400 26 Saturday 25 36 060 400 27 Sunday 24 31 060 406 28 Munday 23 28 060 406 29 Tuesday 22 E 40 060 434 30 Wednesday 22 08 060 434 31 Thursday 21 53 060 444 February 1 Friday 21 40 060 450 2 Saturday 20 47 060 456 3 Sunday 19 22 060 460 4 Munday 19 53 060 472 5 Tuesday 18 23 060 485 D. M. Week days Latitude Dep. E. in leag Dep. W. in leag deg min. 6 Wednesday 17 19 060 518 7 Thursday 16 47 060 556 8 Friday 16 54 060 606 9 Saturday 16 29 060 643 10 Sunday 16 19 060 678 11 Munday 16 21 060 726 12 Tuesday 16 28 060 784 13 Wednesday 16 35 060 840 14 Thursday 16 27 060 889 15 Friday 16 24 060 929 16 Saturday 16 2 060 973 17 Sunday 16 44 060 1003 guess is the Log-line every 2 hours your course and distance is set
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
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