The mariners magazine, or, Sturmy's mathematical and practical arts containing the description and use of the scale of scales, it being a mathematical ruler, that resolves most mathematical conclusions, and likewise the making and use of the crostaff, quadrant, and the quadrat, nocturnals, and other most useful instruments for all artists and navigators : the art of navigation, resolved geometrically, instrumentally, and by calculation, and by that late excellent invention of logarithms, in the three principal kinds of sailing : with new tables of the longitude and latitude of the most eminent places ... : together with a discourse of the practick part of navigation ..., a new way of surveying land ..., the art of gauging all sorts of vessels ..., the art of dialling by a gnomical scale ... : whereunto is annexed, an abridgment of the penalties and forfeitures, by acts of parliaments appointed, relating to the customs and navigation : also a compendium of fortification, both geometrically and instrumentally / by Capt. Samuel Sturmy.

CHAP. II. Containing the Doctrine of the Dimensions of Right-Lined Triangles, whether Right-Angled or Oblique-Angled; and the several Cases therein resolved, both by Tables, and also by the Lines of Artificial Num∣bers, Sines, and Tangents.

I Come now to shew you how a Plain Triangle may be resolved; that is, by ha∣ving any three of the six Parts of a Plain Triangle, to find a fourth by the Instru∣ments before-mentioned.

In all the Cases following I have made use of but two Triangles for Examples; one Right-Angled, the other Oblique-Angled: But in either of them I have expressed all the Varieties that are necessary; so that any three Parts being given in any of them, a fourth may be found at pleasure.

The Sides of any Plain Triangle may be measured by any Measure or Scale of Equal Parts; as an Inch divided into 10 Parts, or 20, 30 Parts; or likewise into Inches, Feet, Yards, Poles, Miles, or Leagues.

[illustration] geometrical diagram

Draw a Line at pleasure, as AB;* 1.1 and from the Point. A let it be requi∣red to protract an Angle of 41 deg. 24 min. First extend the Compasses upon the Line of Chords, from the beginning thereof to 60 deg. always, and with this distance setting one Foot upon the Point A, with the other describe the pricked Arch BC: Then with your Compasses take 41 deg. 24 min. (which is the Quantity of the inquired Angle) out of the Line of Chords, from the beginning

thereof to 41 deg. 24 min. Keeping the Compasses at this distance, if you set one Foot thereof upon B, the other will reach upon the Arch to C. Lastly, draw the Line AC. So the Angle CAB shall contain 41 deg. 24 min.

* 1.2Suppose CAB were an Angle given, and that it were required to find the Quan∣tity thereof. Open your Compasses, as before, to 60 deg. of your Chord; and placing one Foot in A, with the other describe the Arch BC. Then take in your Compasses the Distance CB, and measuring that Extent upon the Line of Chords, from the beginning thereof, you shall find it reach 41 deg. 24 min. which is the Quantity of the inquired Angle.

If any Angle given or required shall contain above 90 Degrees, you must then protract it at twice, by taking first the whole Line, and then the Remainder.

The several Cases of Right-Angled Triangles, may not only be applied to Navigation, but also in the taking of Heights, as is shewn elsewhere: And the Oblique-Angled Triangle, for the taking of Distances, taught in this following Treatise.

In the resolving of Plain Triangles there are several Cases, of which I will only insist on these, which have most relation to the Work in hand.

Of Right-Angled Plain Triangles.

CASE I. In a Right-Angled Plain Triangle, The Base and the Angle at the Base being given, To find the Perpendicular.

SUppose that the Line CA (in the following Figure) in the Right-Angled Trian∣gle, were a Tree, Tower, or Steeple, and that you would know the Height thereof; you must observe with your Instrument the Angle CBA, and measure the Distance BA.

So have you in the Right-Angled Triangle ABC, the Base 405 Foot (Miles or Leagues the denomination might have been as well) and the Angle at the Base 32 deg. and it is required to find the Perpendicular AC.

Now because the Angle CBA is given, the Angle BCA is also given, it being the Complement of the other to 90 deg. and therefore the Angle BCA is 58 Degrees: Then to find the Perpendicular CA, the Proportion is,

As the Sine of the Angle BCA 58 deg. (which is)	9928420
Is to the Logarithm of the Side BA 405 Foot	2607455
So is the Sine of the Angle CBA 32 deg. (which is)	9724210
The Sum of the Second and Third added	12331665
The first Number substracted from the Sum	9928420
To the Logarithm of the Side CA	2403245

The nearest Absolute Number answering to this Logarithm 2403245, is 253 fere; and that is the Length of the Side CA in Miles or Leagues, or the Height of the Tree, Tower, or Steeple, which was required.

A GENERAL RULE.

IN all Proportions wrought by Sines and Logarithms, you must observe this for a General Rule, (viz.) To add the second and third Numbers together, and from the Sum of them to substract the first Number; so shall the Remainder answer your Question demanded, As by the former Work you may perceive, where the Loga∣rithm of the Side BA 2607455 (which is the second Term) is added to the Sine of the Angle CBA 9724210 (which is the third Term) and from the Sum of them, namely from 12331665, is substracted 9928420, the Sine of the Angle BCA,

which is the first Number, and there remaineth 2, 403245, which is the Logarithm of 253 almost, and that is the Length of the Side required.

[illustration] geometrical diagram

To resolve the same Work by the Line of Sines and Numbers.

YOu may work these Proportions more easily by help of the Line of Sines,* 1.3 Tan∣gents, and Numbers, on your Scale, the Proportion being as before.

Therefore if you set one Foot of your Compasses at 58 deg. in the Line of Sines, and extend the other Foot to 405 in the Line of Numbers, the same will reach from the Sine of 32 deg. to 253 in the Line of Numbers, which is the Length of the Side AC, which was required.

Or otherwise, Extend the Compasses from the Sine of 32 deg. to the Sine of 58 deg. in the Line of Sines; the same Extent will reach from 405 in the Line of Num∣bers, to 253, as before, the Work is much abbreviated, there being no need of Pen, Ink, nor Paper, or Tables; but only of your Compasses.

CASE II. The Base and the Angle at the Base being given, To find the Hypothenusa.

IN the same Triangle ABC, Let there be given (as before) the Base AB 405 Foot, Miles, Leagues, or Perches, and the Angle ABC 32 deg. and let it be required to find the Hypothenusa BC. Now because the Angle CBA is given, the other Angle BCA is also given; and the Proportion is,

As the Sine of the Angle BCA 58 deg.	9, 928420
To the Logarithm of the Side 405 Foot	2, 607455
So is the Sine of the Angle CAB 90 deg.	10, 000000
The Sum of the second and third Number	12, 667455 added.
To the Logarithm of the Side BC, which is	2, 679035

The Absolute Number answering to this Logarithm is 478; and so many Feet, Miles, Leagues, Perches is the Hypothenusa, according to the denomination of the Question; that is, whether it be Feet, Perches, Miles, or Leagues. By either of these the Work is the same way.

By the Line of Numbers and Sines.

AS was said before, the Work is altogether the same with the Tables; For the Proportion being,

As the Sine of the Angle BCA 58 degrees

Is to the Length of the Side BA 405 Foot:

So is the Sine of the Angle CAB 90 Degrees,

To the Length of the Side CB 478.

Extend the Compasses from the Sine of 58 deg. to 405 in the Line of Numbers; the same Extent will reach from the Sine of 90 deg. to 478 in the Line of Numbers, and that is the Length of the Side BC. Or you may extend the Compasses from the Sine of 58 deg. to 90 deg. the same Extent will reach 405 to 478, as before.

CASE III. The Hypothenusal and Angle at the Base being given, To find the Perpendicular.

IN the same Triangle let there be given the Hypothenusal BC 478 Feet, Poles, Miles, Leagues, and the Angle at the Base CBA 32 deg. To find the Perpen∣dicular CA.

The Angle CAB is a Right Angle, or 90 Degrees; Therefore the Proportion is,

As the Sine of the Angle CAB 90 deg	10, 000000
Is to the Logarithm of the Side BC 478	2, 679428
So is the Sine of the Angle CBA 32 deg.	9, 724210
To the Logarithm of the Side AC 253	12, 403638

The Number answering to this Logarithm is 253 fere; and that is the Length of the Side CA in Feet, Poles, Miles, or Leagues.

Here the Work is something abbreviated; for the Angle CAB being a Right Angle, and being the first Term, when the second and third Terms are added together, the first is easily substracted from it, by cancelling the Figure next your left hand, as you see in the Example; and so the rest of that Number is the Logarithm of the Num∣ber sought.

By the Line of Sines and Numbers.

EXtend the Compasses from the Sine of 90 Degrees, to 478; the same Extent will reach from the Sine of 32 Degrees, to 253.

Or, Extend the Compasses from the Sine of 90 Degrees, to the Sine of 32 Degrees; the same Extent will reach from 478, to 253; and that is the S••le CA.

CASE IV. The Hypothenusal and Angle at the Base being given, To find the Base.

LEt there be given in the Triangle the Hypothenusal BC, and the Angle at the Base CBA; and by consequence the Angle BCA the Complement of the other to 90 degrees: Then to find BA, the Proportion is,

[illustration]

As the Sine of 90 deg. CAB	10, 000000
To the Hypothenusal CB 478	2, 679428
So is the Sine of the Angle ACB 58	9, 928420
To the Logarithm of the Base AB	12, 607848

The nearest Number answering to 2, 607848, is the Logarithm of 405: And so many Foot or Poles, or if the Question be Miles or Leagues, is the Base or Parallel of Longitude AB.

Now you see the former Figure is turned, and therefore very fitly may have other Denominations (or Names) So that in the Art of Navigation, it will not be unfit to call one of these Sides the Parallel-Side, as AB, or Side of Longitude, or Meridian Distance; the other the Perpendicular-Side, or the Side of Latitude, as CA; and the Hypothenusal, the Side of Distance CB, and the Arches to lay down from the Chords, as before-directed.

By the Line of Sines and Numbers.

THe Angle given, as before, Extend the Compasses from the Sine of 90 deg. unto 478. the same Extent will reach from the Sine of 58 deg. to 405 in the Line of Numbers.

Or, Extend the Compasses from the Sine of 90 deg. to the Sine of 58 deg. the same Extent will reach from 478 to 405, which is the Length of the Base turned up, or Parallel-Line of Longitude, as before said, AB.

CASE V. Let the Perpendicular be the Difference of Latitude 253 Leagues, and the Angle at C, S. W. b. W. 1 deg. 45 min. Westerly, or 58 deg. Let it be given to find the Hypothenusal or Distance upon the Rhomb.

IF the Perpendicular or Difference of Latitude 253 Leagues AC be given, and the Angle at ACB, S. W. b. W. 1 deg. 45 Westerly, or 58 deg. Then by consequence the Angle ABC, or Complement of the Rhomb is also given; taking the first out of 90 deg. then the Hypothenusal may be found thus.

As the Complement Sine of the Rhomb 32 deg. at B	9, 724210
Is to the Logarithm of the Difference of Latitude 253	12, 403121
So is the Sine of the Angle or Radius 90 deg.	10, 000000
To the Logarithm of the Hypothenusal, or Distance upon the Rhomb or Course sailed 478	2, 678911

Here because the Angle CAB is a Right Angle, or 90 Degrees the Radius, and comes in the third place, I therefore only put an Ʋnity before the second Term, and so substract the first Term, and the Remainder is 2, 678911; the Absolute Number answering thereunto is 478, the Side required.

By the Line of Numbers.

EXtend the Compasses from the Sine of 32 deg. to 253 deg. the same Distance will reach from the Sine of 90 deg. to 478, the Side required.

Or, The Distance between the Sine of 32 deg. and 90 deg. will be equal to the Distance between 253 and 478, and giveth the Side required.

CASE VI. The Hypothenusal or Distance Sailed, and the Perpendicular or Diffe∣rence of Latitude given, To find the Rhomb or Angle ABC.

IN the foregoing Triangle, there is given the Hypothenusal or Distance sailed, CB 478 Leagu••s, and the Perpendicular or 253 Leagues difference of Latitude, and it is required to find the Angle ABC, and by it the Rhomb.

As the Logarithm of the Hypothenusal CB 478 Leagues	2, 679428
Is to the Right Angle or Radius 90 deg. CAB	10, 000000
So is the Logarithm of the Perpendicular 253 CA	2, 403121
To the Complement Sine of the Rhomb, or Sine of the An∣gle ABC 32 deg.	9, 723693

The nearest Number answering to 9, 723693, is the Sine of 32 deg. which de∣ducted from 90 deg. there remains the Angle of the Rhomb 58 deg. or S. W. b. W. 1 deg. 45 VVesterly.

By the Line of Numbers.

EXtend the Compasses from 478, to the Sine of 90; the same Distance will reach from 253, to 32 deg.

Or, Extend the Compasses from 478, to 253; the same Extent will reach from the Sine of 90, to the Sine of 32 deg, which is the inquired Angle ABC, and the Com∣plement of the Rhomb.

CASE VII. The Hypothenusal, and the Parallel of Longitude, and the Radius given, To find the Rhomb or Course Sailed.

IN the foregoing Triangle there is given the Hypothenusal or Distance Sailed, CB 478 Leagues, and the Right Angle CAB 90 deg. the Radius, and the Parallel of Longitude or Base 405 Leagues, to find the Course or Rhomb sailed, or the Angle ACB.

As the Hypothenusal or Distance sailed 478 CB	2, 679428
To the Right Angle CAB Radius or Sine of 90 deg.	10, 000000
So is the Parallel of Longitude, or Base AB 405 Leagues	12, 607455
To the Sine of the Angle of the Rhomb or Course sailed 58 or S. W. b. W. 1 deg. 45 Westerly	9, 928027

By the Lines of Sines and Numbers.

EXtend the Compasses from 478 in the Line of Numbers, to the Sine of 90 deg. the same Extent will reach from 405, to the Sine of 32 deg.

Or, Extend the Compasses from 478 to 405; the same Extent will reach from the Sine of 90, to the Sine of 32 deg. ACB, the Angle of the Rhomb or Course sailed, which was required.

Of Oblique-Angled Plain Triangles.

CASE VIII. Having two Angles, and a Side opposite to one of them, To find the Side opposite to the other.

IN the Triangle QRS, is given the Angle QSR 25 deg. 30 min. and the Angle QRS 45 deg. 20 min. and the Side QS 305 Feet; And it is required to find the Side QR.

Here note, That in Oblique-Angled Plain Triangles, as well as in Right-Angled, the Sides are in proportion one to the other, as the Sines of the Angles opposite to those Sides: Therefore,

As the Sine of the Angle QRS 45 deg. 20 min.	9, 851997
Is to the Logarithm of the Side QS 305	2, 484299
So is the Sine of the Angle QSR 25 deg. 30 min.	9, 633984
The Sum of the second and third Terms	12, 118283
The first Term substracted	9, 851997
To the Logarithm of the Side QR	2, 266286

The nearest Absolute Number answering to this Logarithm is 185; and so many Feet is the Side QR.

By the Line of Sines and Numbers.

THe Line of Sines and Numbers will resolve the Triangle by the same manner of Work, as in the other before. For if you extend the Compasses from the Sine of 45 deg. 20 min. to 305 Foot, the same Distance will reach from 25 deg. 30 min. to 185 Foot, and so much is the Side QR.

Or, Extend the Compasses from the Sine af 45 deg. 20 min. to 25 deg. 30 min. the same Distance will reach from 305 to 185, the Length of the Side inquired.

[illustration] geometrical diagram

In like manner if the Angle RQS, 109 deg. 10 min. and the Angle QRS 45 deg. 20 min. and the Side QS 305 Foot, had been given, and the Side RS required, the manner of Work had been the same: For,

As the Sine of the Angle QRS 45 deg. 20 min.	9, 851997
Is to the Logarithm of the Side QS 305	2, 484299
So is the Sine of RQS 109 deg. 10 min. (or 70 deg. 50 min.)	9, 975233
The Sum of the second and third Terms	12, 459532
The first Term substracted	9, 851997
To the Logarithm of the Side RS 405	2, 607535

The Absolute Number answering to this Logarithm is 406, and so much is the Side RS.

In this Case, because the Angle RQS is more than 90 Degrees, you must there∣fore take the Complement thereof to 180 deg. so 109 deg. 10 min. being taken from 180 deg. there remains 70 deg. 50 min. whose Sine is the same with 109 deg. 10 min. And so you must work with all Angles above 90 Degrees; and so will the Comple∣ment to 180, as before-directed, effect the same thing.

By the Line of Numbers and Sines.

EXtend the Compasses from the Sine of 45 deg. 20 min. to 305 Feet, the same Di∣stance will reach from 70 deg. 50 min. to 405.

Or, The Compasses extended from the Sine of 45 deg. 20 min. to 70 deg. 50 min. the same Extent will reach from 305, to 406 in the Line of Numbers, which is the Side RS required.

CASE IX. Two Sides, and an Angle opposite to one of them being given, To find the Angle opposite to the other.

IN the same Triangle, let there be given the Side QS 305, and QR 185 Feet, together with the Angle QSR 25 deg. 30 min. and let it be required to find the Angle QRS. The Proportion is,

As the Logarithm of the Side QR 185	2, 267172
Is to the Side of the Angle QSR 25 deg. 30 min.	9, 633984
So is the Logarithm of the Side QS 305	2, 484299
The Sum of the second and third Numbers	12, 118283
The first Number substracted from the Sum	2, 267172
To the Sine of the Angle QRS 45 deg. 20 min.	9, 851111

The nearest Degree answering to this Sine is 45 deg. 20 min. which is the Angle re∣quired QRS.

By the Line of Sines and Numbers.

EXtend the Compasses from 185, to 25 deg. 30 min. the same Distance will reach from 305, to 45 deg. 20 min. the Angle QRS.

Or, Extend the Compasses from 185, to 305; the same Extent will reach from 25 deg. 39 min. to 45 deg. 20 min. as before.

CASE X. Having two Sides, and the Angle contained between them given, To find either of the other Angles.

FOr the performance of this Problem, Suppose there were given the Side RS 406, and the Side RQ 185, and the Angle comprehended by them, namely the An∣gle at R, 45 deg. 20 min. and it were required to find either of the other Angles.

First, Take the Sum and Difference of the two Sides given; their Sum is 591, and their Difference is 221. Then knowing, that the three Angles of all Right-lined Triangles, are equal to two Right Angles, or 180 Degrees (by the 17th Theor. of Chap. 3.) Therefore the Angle SQR being 45 deg. 20 min. if you substract this Angle from 180 deg. the Remainder will be 134 deg. 40 min. which is the Sum of the two unknown Angles at Q and S; the half thereof is 67 deg. 20 min.

			deg.	mi.
The Side RS	406 Paces.	Two Right Angles	180	00
The Side RQ	185 Paces.	The Angle at R	45	20
The Sum	591 of the Sides given, RS and RQ.	The Sum of the two opposite Angles	134	40
The Difference	221 of the Sides.	The Half-Sum	67	20

The Sum and Difference of the Sides being thus found, and also the Half-Sum of the two unknown Angles, The Proportion by which you must find the Angles se∣verally is,

As the Logarithm of the Sum of the Sides 591	2771587
Is to the Logarithm of the Difference of the Sides 221	344392
So is the Tangent of the Half-Sum of the two unknown Angles 67 deg. 20 min.	10379213
The Sum of the second and third Number	12723605
The first Number substracted from the Sum	2771587
The Tangent of 41 deg. 50 min. (is this)	9952018
Which added to the Half-Sum, makes	109 deg. 10 min. Greater Angle.
The greater of the Angles required, Substract 41 deg. 50 min. from the Half-Sum, leaves the lesser Angle at S	25 deg. 30 min. Lesser Angle.

* 1.4

[illustration] geometrical diagram

By the Line of Sines and Numbers.

EXtend the Compasses from the Sum of the Sides 591, to the Difference of the Sides 221; the same Extent upon the Line of Tangents, will reach from the Half-Sum, to the Tangent of the found Angle 41 deg. 50 min.

Or else extend the Compasses from the Difference 221, to the Tangent of the Half-Sum of the unknown Angles; the same Distance will reach from the Half-Sum 67 deg. 20 m. in the same Line, to the Tangent of 41 deg. 50 min. which added to, or substracted from the Half-Sum, as before is shewn, will give the Quantity of either of the two unknown Angles.

CASE XI. Two Sides and their Containing Angle given, To find the third Side.

THere is given RS 406 Paces, and RQ 185 Paces, and the Angle at R 45 deg. 20 min. which is by the 10 Case,

As the Sum of the Sides given RS + RQ 594 p.	2771587
Is in proportion to their Difference RS + RQ 221	2344392
So is the Tangent of the Half Sum of the 67 deg. 20 min.	10••79213
Two opposite Angles Q and S unknown, 2, 3 Numb.	1••723605
To the Tangent of the Angle	41 d.	50 m.	9952018
Which added to the Half Sum	67	20
Leaves the greater Angle at Q required	109	10
Whose Complement to 180 deg. is	70	50

Then say,

As the Sine of the Angle found 109, or 70 deg. 50 min.	9,975233
Is in proportion to his opposite Side RS 406 Paces	2,608526
So the Sine of the Angle given at R 45 deg. 20 min.	9,851997
To his opposite Side required QS 305 Paces	12,464523
The Logarithm of the Side required	2, 485290

By the Line of Sines and Numbers.

EXtend the Compasses from the Sine of 70 deg. 50 min. to the Logarithm-Side RS 406 Paces; the same Extent will reach from the Sine of 45 deg. 20 min. to the Side 305.

Or, Extend the Compasses from the Sine of 70 deg. 50 min. to the Sine of 45 deg. 20 min. the same Distance will reach from 406, to 305 Paces, which is the Length of the Side QS, which is required.

CASE XII. Three Sides of an Oblique Triangle being given, To find the Angles.

[illustration] geometrical diagram

IN this Triangle SQR, Let the three Sides known,

The Side SR	406
The Side SQ	305
The Side QR	185

And it is required to find the three Angles.

To perform this, you must first let fall a Perpendicular from the Point Q, upon the Side SR, which you may do by setting one Foot of your Compasses in the Point Q, and open the other to the Point R, draw the Arch RE, and divide the Space ER into two equal parts; and so the Perpendicular will fall upon the Point B.

To perform this more exactly by Numbers,

As the greater Side or Base SR, 406	2,608526
To the Sum of the two lesser Sides 490	2,690196
So is the Difference of these two Sides 120	2,079181
To the Part SE (cut off by the Arch RBE) 145	4,769377
	2,608526
	2,161851

This substracted from the whole Line 406, leaves for the part within the Arch 261; the half thereof is 130 ½, which is the Place B where the Perpendicular will fall, reckoned from the Angle R; and by this Perpendicular you have divided the Triangle into two Right Angles, whose Sides are known: For RB being 103 ½, sub∣stracted from the whole Line SR 406, leaves for the remaining Part 275 ½. Now having those two Sides of these two Right-Angled Triangles, and the two first given Sides, 305 and 185, being the two Hypothenusals thereof, you may by the opposition of Sides to their Angles, as in the 6 Case; or by the Sides and Hypothenusal, as in the 7 Case, find the Angles.

By the Line of Sines and Numbers.

EXtend the Compasses from 406, to 490; the same Distance will reach from 120, to 145 SE Leagues, the Side required.

These are the most needful Cases in the Resolution of Plain Triangles, which might have been set forth with much Variety and Inlargement; but I rather strive to shew the best and plainest way. The Practitioner being perfect in what hath been said be∣fore, we will proceed to our intended Discourse of NAVIGATION.

Notes

• * 1.1

  How to lay down an Angle by the Line of Chords.
• * 1.2

  To find the Degrees contained in an An∣gle.
• * 1.3

  Place here the Line of Numbers, Sines, and Tangents.
• * 1.4

  ½ Sum	67	20
  Tang.	41	50
  	109	10
  Lesser Angle	25	30