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.

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.

* 1.1

[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

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