Thesaurarium mathematicae, or, The treasury of mathematicks containing variety of usefull practices in arithmetick, geometry, trigonometry, astronomy, geography, navigation and surveying ... to which is annexed a table of 10000 logarithms, log-sines, and log-tangents / by John Taylor.

CHAP. V. Of TRIGONOMETRY. Or the Doctrine of Triangles.

SECT. I. Some general Maxims, belonging to plain or Right-lined Triangles.

TRIGONOMETRY is necessary in most parts of the Mathematicks, and herein indeed consisteth the most fre∣quent use of the Logarithms, Sines, Tangents, and Secants: It is conversant in the measuring of Triangles, Plain or Spherical, comparing their Sides, and Angles together; according unto their known Analogies, or Proportions: So that any three parts of a Triangle being given, the other parts may be found out, and known: Now in the Doctrine of Right-lined Triangles, it will be necessary to know these Maxims fol∣lowing.

1. That a Right-lined Triangle, is a Figure constituted, by the Conjunction, or Intersection, of the three Right, or Streight-lines thereof; in their Angles or Meeting-places. So that e∣very Triangle hath six distinct parts, Viz. Three Sides, and three Angles.

2. That all Right-lined Triangles, are either* 1.1 Right-angled, That is, which hath one Right∣Angle, as ABC Fig. 34. Or Oblique-angled, whose three Angles are all Acute; that is, less than a Quadrant, or 90 deg; or else they have One Angle Obtuse, or greater than a Quadrent: So all Triangles, that have not one Right-angle,* 1.2 are called Oblique-Triangles; as Fig. 36. to wit, the Triangle ABC.

3. That the three Angles, of any Right-lined Triangle, are equal unto two Right-angles; or 180 Degrees. So that any two of their Angles being known, the third Angle is also found, being the Complement of the other two; unto 180 Degrees: But this is more readily found in a Rectangled Triangle, for the Rectangle being a Quadrent, or 90 degrees, one of the acute Angles therefore being given, the other is rea∣dily known, being the Complement thereof unto 90 Degrees.

4 That the three sides, comprehending the* 1.3 Triangle, some call Leggs, others Sides, but in Rectangled Triangles, as in the Triangle ABC, I call AB, the Base, BC the Cathetus or Perpendi∣cular; and AC the Hypothenuse.

5. That the Sines, of the Angles are proporti∣onal unto their opposite Sides; and their Sides, to their opposite Angles. So that if the Side of a Triangle were desired, put the Sine of the oppo∣site

Angle in the first place. Also if an Angle be required, put the Logarithm of his opposite side in the first place.

6. That the sides of any Rectangled Tri∣angle may be measured by any Scale of equal parts, as Inches, Feet, Yards, Poles, Miles, Leagues, &c.

7. That if an Angle propounded, be greater than 90 deg. and so not to be found in the Ta∣bles, take the Complement thereof, unto 180 deg. and work by the Sine, or Tangent there∣of, and the work will be the same.

And here for the more short, and speedy performance of these conclusions in Trigono∣metry; I have annexed, and used, these fol∣lowing Symbols; which I would have you take notice of.

• = Equal, or Equal to.
• + More.
• - Less.
• × Multiplyed by.
• ° Degrees as 15°.
• ' Minutes as. 40'.
• cr. A Side.
• cr^s, Sides.
• V An Angle.
• VV Angles.
• Z Sum.
• X Difference.
• S Sine.
• Sc Co-sine.
• T Tangent.
• Tc Co-tangent.
• Se Secant.
• Sec Co-secant.
• Co. Ar. Compl. Arithmetic.
• R A Right-angle.
• 2R Two Right-angles.
• Q Square.

SECT. II. Of Plain Rectangled Triangles.

PROP. I. Two Angles and the Base of a Rectangled Trian∣gle given, to find the other parts.

ADmit the Triangle given be ABC: Now* 1.4 the Angle at B, is an Angle of 90°, or a Right-angle; And the Angle at C is 57° 35', and the Base AB; is 736 parts.

Now first I find the Angle at A, to be 32° 25': it being the Complement of the Angle at C, unto 90°: Secondly, to find the Cathe∣tus, or Perpendicular, this is the analogy or proportion.* 1.5

Add the Log. of the third and second Terms together, and from their Sum, deduct the Log. of* 1.6 the first number, so is the Remainder, the Log. of

the fourth Term, or Number sought, as you see in the aforegoing Example.

Thirdly to find the Hypothenuse AC, the Analogy or Proportion hold thus.

As S. V, C 57° 35',

To Log. Base AB 736 ••arts.* 1.7

So Radius or S. 90°,

To Log. Hypothenuse AC 871 8/10 parts re∣quired: Thus are the three required parts, of the given Triangle ABC found, viz. the Angle A to be 32° 25', the Cathetus BC to be 467 4/10 parts, and the Hypothenuse AC to be 871 8/10 parts, as was so required to be found.

PROP. II. The Hypothenuse, Base, and one of the Angles Of a Rectangled Triangle given, to find the o∣ther parts thereof.

In the Triangle ABC, the Hypothenuse AC is 871 8/10 parts, the Base AB is 736 parts, and the Angle at B, is known to be a Right-angle; or 90°: First to find the Angle at the Cathetus C, the analogy or proportion holds thus.

As Log. Hypothen: AC 871 8/10 parts* 1.8

To Radius or S. 90°.

So Log. Base AB 736 parts,

To the S. V. at Cathetus C 57° 35'.

Secondly, now having found the Angle at the Cathetus C, to be 57° 35'; I say the Angle of the Base A is 32° 25', being the Compl. of the Angle C, unto 90°.

Thirdly to find the Cathetus BC, this is the ••nalogy, or proportion.

As Radius or S. 90°,

To Log. Hypothen. AC 871 8/10 parts,

So S. V. at Base A 32° 25'.* 1.9

To Log. Cathetus BC, 467 4/10 parts required. It may also be found, as in the former Proposi∣tion.

PROP. III. In a Rectangled Triangle, the Base, and Cathetus given to find the other parts thereof.

In the Triangle ABC, the Base AB is 736 parts, and the Cathetus BC is 467 4/10 parts, and the Angle B, between them is a right angle o•• 90°: And here you may make either side of the Triangle, Radius, but I shall make BC the Cathetus Radius, and then to find the Angle at the Cathetus C, this is the Analogy or ••••••∣portion.

As Log. Cathet, BC 467 4/10 parts,* 1.10

To Radius or S 90°.

So Log. Base AB 736 parts,

To T. V. Cathe C 57° 35', as required.

Secondly, I find the other Angle, at A to be* 1.11 32° 25', it being the Complement, to C 57 35', unto 90°.

Thirdly, To find out the Hypothenuse AC this is the analogy or proportion.

As S. V. Cathe C. 57° 35',

To Log. Base AB 736 parts,

So Radius or S. 90°,

To Log. Hypothenuse AC 871 8/10 parts, 〈…〉〈…〉∣quired. But making the Base AB Radius, yo•• may find the Hypothenuse AC, by this anal〈…〉〈…〉 or proportion.

[illustration]

Plate 1 Page 65

As Radius or S. 90°,

To Log. Base AB 736 parts.

So Sc. V. Base A 32° 25',

To Log. Hypothenuse AE 871 8/10 parts required,* 1.12 and thus you have all the parts of the Trian∣gle propounded.

PROP. IV. The Base, and Hypothenuse, with the Angle be∣tween them given, to find the other parts of a Rect-angled Triangle.

In the Triangle ABC, the Base AB is 736 parts, and the Hypothenuse AC is 871 8/10 parts, and the Angle A included between them is 32° 25'. First to find the Angles, and first remem∣ber that the Angle B is a right Angle; or 90°. Secondly, that the Angle at C, is the Comple∣ment to the Angle at A 32° 25' unto 90°:* 1.13 and therefore is 57° 35': Now these being known, you may find the Cathetus, by this analogy or proportion.

As S. V. Cathe. C. 57° 35',

To Log. Base AB 736 parts.

So S. V. Base A 32° 25',

To Log. Cathe. BC 467 4/10 parts required. Thus I have sufficiently explained all the Cases of Plain Rect-angle Triangles, for to these rules they may be all reduced.

SECT. III. Of Oblique-Angled Plain Triangles.

PROP. I. Two Angles, and a side opposite, in an Oblique-Ang∣led Triangle given, to find the other parts there∣of.

IN the Triangle ABC, the Angleat A is 50°, and at C is 37°, and the side AB is 30 parts, and opposite to the Angle C

First, to find the Angle B, remember that (as 'tis said, in the third Maxim aforegoing) 'tis the Complement, to the Angles A 50°, and C 37°, to 180°, and therefore is the Angle at B 93°.

Secondly, having thus found the Angles, the* 1.14 two unknown sides, may be found by the pro∣portion they bear to their opposite Angles, for that proportion holds also in these; thus to find the side BC, this is the analogy or proportion.

As S. V. C 37° 00',

To Log. side AB 30 parts.

So S. V. A. 50° 00',

To Log. side BC 38 19/100 parts required to be found.

But it may be more readily found, and per∣formed in such case as this, where you have a Sine, or Tangent, in the first place, by the A∣rithmetical Complement thereof, and so save the Substraction.

Now the readiest way to find the Arithmetical* 1.15 Complement is that of Mr. Norwood, in his Doctrine of Triangles; which is thus: begin with the first Figure towards the left hand of any Number and write down the Complement, or the re∣mainder thereof, unto 9:* 1.16 And so do with all the* 1.17 rest of the Figures, as you see here done. Saying 9, wants of 9, 0: and again 9, wants 0: 6, wants 3; 2, wants 7: 3, wants 6; 9, wants 0: only when you come to the last Figure to the right hand, take it out of 10, so 8, wants 2; of 10: Thus you may readily find the Co-Ar. of any Sine, al∣most as soon as the Sine it self.

But if you want the Complement Arithmetical of any Tangent, you may take the Co-tang. which is exactly the Co-Arith. of the double Radius, so that the Tangent, and Co-tangent, of an Arch makes exactly 20. 000000.

Now if the Radius be in the first place, then there is no need of taking the Co-Arith. of the first Number, only you must cut off, the first I, to the left hand thus X, and you will have the Logarithm of the Number desired.

Thirdly, now to find the side AC, by the opposite Angle B; which is 93° 00': And see∣〈…〉〈…〉ng the Angle B, exceeds 90°, you must work 〈…〉〈…〉y the Complement to 180°) as in the seventh 〈…〉〈…〉ork in page 61 is taught.

Thus having found all the parts of the Tri∣angle * 1.18 * 1.19 propounded, Viz. The Angle B, to be 93° 00', the side AC to be 49 78/100 parts, and the side BC to be 38 19/100 parts, as was required to be found.

PROP. II. Two sides, and an Angle opposite to one of them in an Oblique-angled Triangle given, to find the other parts thereof.

In the Triangle ABC, the side AB is 30 parts, and the side AC, is 49 78/100 parts, and the opposite Angle C, is 37° 00'.

First, To find the Angle at B, this is the A∣nalogy* 1.20 or Proportion.

As Log. cr. AB 30 parts,

To S. V. at C 37° 00'.

So Log. cr. AC 49 78/100 parts,

To Sc. V. B 93° 00', as was required to be found.

Now seeing that the Angle C, is 37° 00', and the Angle B, is 93° 00', which makes 120° 00', therefore must the Angle A be 50° 00'; the

Complement to 180°: so having found all the three Angles, you may find the other side CB,* 1.21 38 19/100 parts, as afore in the first proposition, by his opposite Angle.

PROP. III. Two Sides of an Oblique-angled Triangle, with the Angle included between them given, to find the other parts thereof.

In the Triangle ABC, the side AC is 49 78/100 parts, the side DB is 30 parts, and the Angle A between them is 50° 00'; and 'tis required to find the other parts of the Triangle propoun∣ded. To resolve this Conclusion, let fall a Perpendicular DB, from the Angle B, on the side AC; (by prop. 3. §. 1. chap. 4) and then pro∣ceed thus.

First, Seeing the Oblique-angled Triangle,* 1.22 ABC is divided into two Rectangled Triangles, Viz. ADB, and BDC: Now I will begin with the Triangle ADB, in which is given the An∣gle A 50° 00', and the Angle D is a right An∣gle, or 90°, and the side AB 30 parts, and the sides AD, and DB, and the Angle at B, are required.

First to find the Angle at B, remember that it is the Complement unto the Angle A 50° 00', unto 90° 00', and therefore must the Angle B be 40° 00'; Now for to find the Cathetus BD, (as in prop. 1. and 2 §. 2. chap. 5.) by the Rule of opposition, the Analogy or Proportion holds thus.

As Radius or S. 90°,

To Log. Hypoth. AB 30 parts.

So S. V. at A 50° 00',

To Log. Cath. BD 22 98/100 parts sought.

And AGAIN, say.

As Radius or S. 90°,

To Log. Hypoth. AB 30 parts.

So S. V. at B 40° 00',

To Log. Base AD 19 28/100 parts sought.

Thus in the Triangle ADB, you have found the Angle B, to be 40° 00', the Cathetus BD, to be 22 98/100 parts; and the Base AD to be 19 28/100 parts, as was so required.

Now for the other Triangle which is BDC, in which there is given the side BD, 22 98/100 parts, and the Angle at D, is a Right-angle, or 90°, and the sides DC, and CB, and the Angles B, and C, are required.

First to find the side DC, substract AD, 19 28/100* 1.23 parts, out of AC, 49 78/100 parts; there remains the Base DC; 30 50/100 parts: Thus have you the two sides of the Triangle, to wit the Base DC, 30 50/100 parts, and the Cathetus BD, 22 98/100 parts, and the Angle D between them is a Right-angle or 90°. Now you may find the Angle at B, by the Tangent (as in prop. 3. §. 2. chap. 5.) thus.

As Log. Cath. BD, 22 98/100 parts,

To Radius or S. 90°.

So Log. Base CD 30 50/100 parts.

To T. V. B. 53° 00'.

Secondly, For the Angle C, remember 'tis the Complement of the Angle B, 53°, to 90°; and therefore is the Angle C, 37° 00', required.

Thirdly, To find the Hypoth. BC, this is the Analogy or Proportion.

As S. V. B. 53° 00',

To Log. Base DC 30 50/100 parts.

So Radius or S. 90°,

To Log. Hypoth. BC 38 19/100 parts: Thus have you found all the required parts of the Tri∣angle ABC propounded, viz. the Angle C to be 37° 00', the Angle B, to be 93° 00', * 1.24 and the Side BC, 38 19/10•• parts, as required to be found.

Another way to perform the same.

Take the Sum of the* 1.25 two sides, and the diffe∣rence of the two sides; and work as followeth.

Now to find the two* 1.26 Angles B, and C, this is the Manner, and by this Analogy or Proportion, they are found out and known.

As Log. Z. cr^s. AB, and CA, 79 78/100 parts,

To Log. X. cr^s. AB, and CA; 19 78/100 parts,

So T. of ½ VV unknown, 65° 00',

To T. ½X. of VV, 28° 00'.

This difference of Angles 28° 00', add unto 65° 00', (half the difference of the unknown Angles) and it shall produce 93° 00', which is the greater Angle, and substracted from it, leaves 37° 00', which is the lesser Angle C: so have you the required Angles.

PROP. IV. The three sides of an Oblique-angled Triangle gi∣ven, to find the Angles.

In the Triangle ABC, the side AC, is 49 78/100 parts, the side AB, is 30 parts, and the side BC, is 38 19/100 parts; and the three Angles of the Tri∣angle are required.

The resolution of* 1.27 this Conclusion is thus. Take the Summ and Dif∣fer.* 1.28 of the two sides AB, and BC; And then work as follows: To find a Segment of the Base AC, to wit CE; say:

As Log. Base AC, 49 78/100 parts,

To Z. cr^s. AB, and BC; 68 19/100 parts,

So X. cr^s. AB, and BC; 8 19/100 parts,

To Log of a Segment of the Base AC, to wit C E 11 22/100 parts.

This Segment of the Base CE, 11 22/100 parts, being substracted from the whole Base AC, 49 78/100 parts, the remainder is EA 38 56/100 parts, in the middle of which as at D, the Perpendicu∣lar DB, will fall from the Angle B; and so di∣vide* 1.29 it into two Rectangled Triangles, to wit, ADB, and CDB, whose Base DA is 19 28/100 parts, which taken from AC 49 78/100 parts, leaves the Base of the greater Triangle CD 30 50/100 parts.

Now having the two Bases of these two Tri∣angles, and their Hypothenuses; to wit CD 30 50/100 parts, DA 19 28/100 parts, CB 38 19/100 parts, and BA 30

parts; you may find all their Angles, by the Rule of Opposite sides, to their Angles as afore.

I. In the Triangle CDB.

To find the Angles, this is the Analogy or Proportion.

As Log. BC 38 19/100 parts,

To Radius or S. 90°.

So Log. DC 30 50/100 parts,

To S. V. B 53° 00': whose Complement is the* 1.30 Angle at C 37° 00' unto 90: or a Quadrant.

II. In the Triangle ADB.

To find the Angles, this is the Analogy or Proportion.

As Log. AB, 30 parts,

To Radius or S. 90°.

So Log. AD 19 28/100 parts,

To S. V. B, 40° 00'.

The Complement whereof, unto 90° 00', is the Angle at A 50° 00'.

Now in the first Triangle CDB, there is found the Angle C, to be 37° 00', and the Angle B, to be 53° 00'.

In the second Triangle ADB, there is found the Angle A; to be 50° 00', and the Angle B, to be 40° 00'.

Now the two Angles at B, to wit 53° 00'; and 40° 00'; makes 93° 00', which is the Angle of the Oblique-angled Triangle ABC, at B: Thus the three Angles of the said given Triangle ABC, are found as was required, viz. the An∣gle A to be 50° 00', the Angle B to be 93° 00', and the Angle C to be 37° 00', as sought.

Thus I have sufficiently, fully and plainly explained all the Cases of Plain Right-lined Tri∣angles, both Right and Oblique-angled: I shall now fall in hand with Spherical Triangles, both Right and Oblique-angled.

SECT. IV. Of Spherical Rectangled Triangles.

And here first it will be necessary also to understand those few general Maxims or Rules, that are of special Moment, in the Doctrine of Spherical Triangles.

1. THat a Spherical Triangle is comprehen∣ded and formed, by the Conjunction and Intersection of three Arches of a Circle, described on the Surface of the Sphere or Globe.

2. That those Spherical Triangles, consisteth of six distinct parts, viz. three Sides and three Angles, any of which being known, the other is also found out and known.

3. That the three Sides of a Spherical Trian∣gle, are parts or Arches of three great Circles of a Sphere, mutually intersection each other: and as plain or Right-lined Triangles, are mea∣sured by a Measure, or Scale of equal parts: So these are measured, by a Scale or Arch of equal Deg••ees.

4. That a Great Circle is such a Circle that doth bessect the Sphere, dividing it into two equal parts; as the Equinoctial, the Ecliptick, the Meridians, the Horizon, &c.

5. That in a Right-angled Spherical Triangle, the Side subtending the Right-angle we call the Hypothenuse, the other two containing the Right∣angle we may simply call the Sides, and for distinction either of them may be called the Base or Perpendicular.

6. That the Summ of the Sides of a Spherical Triangle are less than two Semicircles or 360°.

7. That if two Sides of a Spherical Triangle be equal to a Semicircle; then the two Angles at the Base shall be equal to two Right-angles; but if they be less, then the two Angles shall be less; but if greater, then shall the two Angles be greater than a Semicircle.

8. That the Summ of the Angles of a Spheri∣cal Triangle, is greater than two Right-angles.

9. That every spherical Triangle is either a Right, or Oblique-angled Triangle.

10. That the Sines of the Angles, are in pro∣portion, unto the Sines of their opposite Sides; and the Sines of their opposite Sides, are in proportion unto the Sines of their opposite Angles.

11. That in a Right-angled Spherical Triangle, either of the Oblique-angles, is greater than the Complement of the other, but less than the Diffe∣rence of the same Complement unto a Semicircle.

12. That a Perpendicular is part of the Arch of a great Circle, which, being let fall from any Angle of a spherical Triangle, cutteth the oppo∣site Side of the Triangle at Right-angles, and so

divideth the Triangle into two Right-angled Tri∣angles, and these two parts (either of the Sides or Angles) so divided must be sometimes added together, and sometimes substracted from each other, according as the Perpendicular falls with∣in or without the Triangle.

PROP. I. Case 1. A Side and an Angle adjacent thereunto being gi∣ven, to find the other Side.

In the Triangle ABC, there is given the Side AB 27° 54'; and the Angle A 23° 30', and the Side BC is required, to find which this is the Analogy or Proportion.* 1.31* 1.32

PROP. II. Case 2. A Side and an Angle adjacent thereunto being gi∣ven, to find the other Oblique-angle.

In the Triangle ABC, there is given the Side AB 27° 54', and the Angle A 23° 30',* 1.33 and the Angle at C is required, to find which say by this Analogy or Proportion.

As the Radius or S 90° 00',

To Sc. of cr. AB 27, 54.

So is S. V. at A 23, 30,

To Sc. V. at c 69, 22 required.

PROP. III. Case 3. A Side and an Angle adjacent thereunto being given, to find the Hypothenuse.

In the Triangle ABC, there is given the Side* 1.34 AB 27° 54', and the Angle at A 23° 30', and the Hypothenuse AC, is required; which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To Sc. of V. at A 23, 30.

So is Tc cr. AB, 27, 54.

To Tc. Hypothenuse AC, 30, 00 required.

PROP. IV. Case 4. A Side and an Angle opposite thereunto being given, to find the other Oblique-angle.

In the Triangle ABC, there is given the Side* 1.35 BC 11° 30', and the Angle A 23° 30', and the Angle C is required, to find which, say by this Analogy or Proportion.

As Sc. cr. BC, 11° 30',

To Radius or S. 90, 00.

So is Sc. V. at A, 23, 30,

To S. V. at C. 69, 22, as required.

PROP. V. Case 5. A Side and the opposite Angle given, to find the Hypothenuse.

In the Triangle ABC, there is given the* 1.36 side BC 11° 30', and the Angle at A 23° 30', and the Hypothenuse AC, is required, which may be found by this Analogy or Proportion.

As S. V. at A 23° 30',

To Radius or S. 90, 00.

So is Ser. BC 11. 30,

To S. Hypothenuse AC 30, 00. as required.

PROP. VI. Case 6. A side and the opposite Angle given, to find the other side.

In the Triangle ABC, there is given the side BC 11° 30', and the Angle at A 23° 30', and the side AB is required, to find which this is the Analogy or Proportion.

As Radius or S 90° 00',* 1.37

To Tc. of V. at A. 23. 30,

So is T. cr. BC 11, 30,

To S. of cr. AB 27. 54 as was required.

PROP. VII. Case 7. The Hypothenuse, and an Oblique Angle given, to find the side adjacent thereunto.

In the Triangle ABC, there is given the

Hypothenuse AC, 30° 00', and the Angle A 23° 30', and the side AB, is required, which is found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To Sc. V. at A, 23, 30.* 1.38

So is T. Hypoth. AC, 30, 00,

To T. cr. AB, 27, 54, as was required.

PROP. VIII. Case 8. The Hypothenuse, and an Oblique-angle given, to find the opposite Side.

In the Triangle ABC, there is given the Hypothenuse AC, 30° 00', and the Angle at A 23° 30', and the Side BC, is required, which is found by this Analogy or Proportion.* 1.39

As the Radius or S. 90° 00',

To S. Hypoth. AC, 30, 00.

So is S. V. at A, 23, 30,

To the S. cr. BC, 11, 30. which was required.

PROP. IX. Case 9. The Hypothenuse, and an Oblique-angle given, to find the other Oblique-angle.

In the Triangle ABC, there is given the Hy∣pothenuse AC 30° 00', and the Angle A, 23° 30', now the Angle at C, is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To Sc. Hypoth. AC, 30, 00.

So is T. of V. at A, 23, 30,* 1.40

To Tc. of V. at C. 69, 22, as was required.

PROP. X. Case 10. The sides given, to find the Hypothenuse.

In the Triangle ABC, there is given the side AB 27° 54', and the side BC 11° 30', and the Hypothenuse AC is required, to find which say by this Analogy or Proportion.

As the Radius or S. 90° 00',* 1.41

To Sc. cr. BC. 11, 30.

So is Sc. cr. AB 27, 54,

To Sc. Hypothenuse AC 30, 00. required.

PROP. XI. Case 11. The sides given, to find an Angle.

In the Triangle ABC, there is given, the side AB 27° 54', and the side BC 11° 30', and the Angle at A, is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',* 1.42

To S. cr. AB. 27, 54.

So is Tc. cr. BC. 11, 30,

To Tc. of V. at A. 23. 30. as required.

PROP. XII. Case 12. The Hypothenuse, and a side given, to find the o∣ther side.

In the Triangle ABC, there is given, the Hy∣pothenuse* 1.43 AC 30° 00', and the side AB 27° 54' and the side BC is required, which may be

found by this Analogy or Proportion.

As Sc. cr. AB. 27° 54',

To Radius or S. 90 00.

So is Sc. Hypothenuse AC. 30° 00',

To Sc. cr. BC. 11° 30' as required.

PROP. XIII. Case 13. The Hypothenuse, and a Side given, to find the contained Angle.

In the Triangle ABC, there is given the Hy∣pothenuse* 1.44 AC 30° 00', and the side AB 27° 54', and the Angle at A is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',

To T. cr. AB. 27° 54'

So is Tc. Hypoth. AC 30° 00',

To Sc. of V. at A, 23° 30', as required.

PROP. XIV. Case 14. The Hypothenuse, and a Side given, to find the oppo∣site Angle.

In the Triangle ABC, there is given the Hy∣pothenuse AC 30° 00', and the side AB 27° 54', ••ow the Angle C, is required, which may be* 1.45 〈…〉〈…〉ound by this Analogy or Proportion.

As the S. Hypoth. C, 30° 00',

To Radius or S. 90° 00'.

So is S. of cr. AB, 27° 54',

To S of V. at C. 69 22, as required.

PROP. XV. Case 15. The Oblique Angles given, to find either Side.

In the Triangle ABC, there is given the An∣gle A 23° 30', and the Angle at C 69° 22', and the side BC, is required, which may be found by this Analogy or Proportion.

As the S. of V. at C, 69° 22',* 1.46

To the Radius or S. 90° 00'.

So is the Sc. of V. at A, 23° 30',

To the Sc. of cr. BC, 11° 30', as required.

PROP. XVI. Case 16. The Oblique-angles given, to find the Hypothenuse.

In the Triangle ABC, there is given the An∣gle A 23° 30', the Angle C, 69° 22', and the Hypothenuse AC, is required, which may be found by this Analogy or Proportion.

As the Radius or S. 90° 00',* 1.47

To Tc. of V. at C. 69°, 22',

So is Tc. of V. at A, 23 30,

To Sc. Hypoth. AC, 30 00, as required.

SECT. V. Of Oblique-angled Spherical Triangles.

PROP. I. Case 1. Two Sides, and an Angle opposite to one of them given, to find the other opposite Angle.

IN the Triangle ADE, there is given the Side* 1.48 AE, 70° 00', the Side ED, 38° 30', and the Angle A, 30° 28', now the Angle at D, is re∣quired, to find which this is the Analogy or Proportion.

As S. cr. DE, 38° 30',

To S. V. at A, 30 28.

So is S. cr. AE, 70 00,

To S. V. at D, 130 03,* 1.49 required.

PROP. II. Case 2. Two Angles and a Side opposite to one of them gi∣ven, to find the Side opposite to the other.

In the Triangle ADE, there is given the An∣gle* 1.50 at D, 130° 03', the Angle E, 31° 34', and the Side AE, 70° 00', now the Side AD, is re∣quired, which may be found by this Analogy or Proportion.

As S. V. at D, 130° 03',

To S. cr. AE, 70 00.

So is S. V. at E, 31 34,

To S. cr. AD. 40 00, required.

PROP. III. Case 3. Two Sides and an Angle included between them being known, to find the other Angles.

In the Triangle ADE, there is given the Side* 1.51 AE, 70° 00', the Side AD, 40° 00', and the An∣gle A 30° 28', Now the Angles D, and E, are re∣quired, which is thus found: take the Sum and Difference of the two Sides, and work as follow∣eth, saying.* 1.52

As S. ½ Z. cr^s. AE and AD, 55° co',

To S. ½ X. cr^s. AE and AD, 15 00.

So is Tc. ½ V. at A, 15 14,* 1.53

To T. ½ X. VV. D and E. 49 1430".

AGAIN.

As Sc. ½ Z. cr^s. AE and AD, 55° 00',

To Sc. ½ X. cr^s. AE and AD, 15° 00'.

So is Tc. ½ V. at A, 15° 14',

To T. ½ Z. VV. D and E, 80 48 30".

This difference of the Angles unknown D and E, 49° 14' 30", being added unto the half Sum of the Angles 80° 48' 30", (unknown) produceth the Greater Angle D 130° 03', and substracted from it, leaves the Lesser Angle E, to wit 31° 34'.

PROP. IV. Case 4. Two Angles, and their Interjacent side being known, to find the other sides.

In the Triangle ADE, there is given the Angl••* 1.54 at A 30° 28', and the Angle at D 130° 03', and their Interjacent-side AD 40° 00', and the Sides DE and EA, are required: Which is thus found.

Take the Sum and Dif∣fference of the two An∣gles,* 1.55 and work as fol∣loweth, saying.

As S. ½ Z. of VV. A and D, 80° 15' 30", * 1.56 * 1.57

To S. ½ X. of VV. A and D, 49 47 30.

So is T. ½ cr. AD, 20 00 00,

To T. ½ X. cr^s. DE and EA, 15 45 00.

AGAIN Say.

As Sc. ½ Z. of VV. A and D, 80° 15' 30",

To Sc. ½ X. of VV. A and D, 49 47 30.

So is T. ½ cr. AD, 20 00 00,

To T. ½ cr^s. Z. DE and AE. 54 15 00.

Add the half Difference of the Sides DE and AE, 15° 45', unto half the Sum of the Sides DE and AE, 54° 15'. It produceth the greater Side, the Side AE 70° 00', but if deducted from it, leaves the lesser Side ED, which is 38° 30', as was required.

PROP. V. Case 5. Two Sides and an Angle opposite to one of them gi∣ven, to find the third side.

In the Triangle ADE, there is given the Side* 1.58 AE 70° 00', the Side DE 38° 30', and the Angle A 30° 28', the Side AD is required.

First by Case 1. Prop. 1. I find the Angle at D to be 130° 03', and then proceed thus

First take the Sum and Difference of the two Angles; then also find the Difference of the two Sides given, and then work as followeth.* 1.59 * 1.60

Now say,

As S. ½ X. VV. D and A, 49° 47' 30",

To S. ½ Z. VV. D and 〈…〉〈…〉, 80 15 30.

So is T. ½ X. cr^s. AE and ED, 15 45 00,

To T. ½ cr. AD. 20° 00' 00": which doubled giveth the Side AD, 40 00 00, as was required.

PROP. VI. Case 6. Two Angles and a Side opposite to one of them gi∣ven, to find the third Angle.

In the Triangle ADE, there is given the An∣gle* 1.61 A 30° 28', the Angle D 130° 03', and his opposite Side AE 70° 00', and 'tis required to find the Angle at E.

First by Prop. 2. Case 2. I find the Side DE, opposed to the Angle A; to be 38° 30', then proceed thus.

Fi••st find the Sum and Difference of the Sides. Then find the Difference of the Angles.* 1.62 * 1.63

Now say,

As S. ½ X. cr^s. DE and AE 15° 45',

To S. ½ Z. cr^s. EA and DE 54 15.

So is T. ½ X. VV. D and A 49 47 30",

To Tc. ½ V. at E 15° 47' 00". which doubled giveth the Angle at E 31 34, as required.

PROP. VII. Case 7. Two Sides and an Angle opposite to one of them gi∣ven, to find the Included Angle.

In the Triangle ADE, there is given the Side* 1.64 AE 70° 00', the Side ED 38° 30', and the An∣gle opposite thereunto at A 30° 28', and the Angle E is required.

First by Prop. 1. Case 1. I find the Angle D, opposite to AE, to be 130° 03', then proceed thus.

First find the Difference of the Angles, then find the Sum and Difference of the Sides.* 1.65 * 1.66

Now say,

As S. ½ X. cr^s. AE and ED 15° 45',

To S ½ Z. cr^s. AE and ED 54 15.

So is T. ½ X. of VV. D and A 49 47 30",* 1.67

To Tc. ½ V. at E 15° 47'. Which doubled is the Angle at E 31° 34', as was required.

PROP. VIII. Case 8. Two Angles and a Side opposite to one of them be∣ing known, to find the Interjacent Side.

In the Triangle ADE, there is given the An∣gle E 31° 34', the Angle D 130° 03', and his opposite Side AE 70° 00', Now the Side ED is required.

First by Prop. 2. Case 2. I find AD opposed to E, to be 40° 00', and then work thus.

Take the Sum and Difference of the Angles, then also find the Difference of the two Sides:* 1.68 * 1.69 * 1.70

Now say,

As S. ½ X. VV D and E 49° 14' 30",

To S. ½ Z VV D and E 80 48 30.

So is T ½ X cr^s. AD and AE 15 00 00",

To T. ½ cr^s. ED, 19° 15' 00", which being doubled is the Side ED 38° 30', as required.

PROP. IX. Case 9. Two Sides and their Included Angle being known, to find the third Side.

In the Triangle APZ, there is given the Side* 1.71 ZP 38° 30', the Side PA 70°, and the Angle P, let be 31° 34, and the Side AZ is required.

The Resolution of this Case depends on the Catholike proposition of the Lord of Marchiston, by supposing the Oblique-Triangle to be divided (by a supposed Perpendicular falling either within or without the Triangle) into two Rect∣angulars.

Now in the Triangle AZP, let fall the Per∣perpendicular ZR; so is the Triangle AZP divi∣ded into two Rectangulars ARZ and ZRP. Now the Side AZ may be found at two Opera∣tions thus: say,

As the Radius or S. of 90° 00'

To Sc. of the included V, P. 31 34.

So is T. of the lesser Side PZ. 38 30,

To T. of a fourth Arch. 34 08.

If the contained Angle be less than 90°, take this fourth Arch from the greater Side; but if it be greater than 90°, from its Complement unto 180°, the Remainder is the Residual Arch: Now again say,

As Sc. of the fourth Arch. 34° 08'

To Sc. Residual Arch. 35 52

So Sc. of the lesser Side PZ. 38 30

To Sc. AZ the Side required. 40 00

☞ 1.72 But note that many times the Perpendicu∣lar will fall without the Triangle, as it doth* 1.73 now within; in such case the Sides of the Tri∣angle must be continued, so will there be two Rectangulars, the one included within the o∣ther: as in the Triangle HIK, the Perpendicular let fall is KM, falling on the Side HE, and so the two Rectangulars found thereby will be IM K, and KMH, and so by the directions in the former proposition find out the Side IK, if re∣quired to be found.

PROP. X. Case 10. Two Angles and their Interjacent Side known, to find the third Angle.

In the Triangle AZP, there is given the Side ZP 38° 30', the Angle P 31° 34', and the An∣gle Z 130° 03', and the Angle at A is requi∣red.

First the Oblique-Triangle AZP, being redu∣ced into two Rectangulars ARZ, and ZRP, by* 1.74 Case 9 aforegoing, I find the Angle RZP, to be 64° 19', (in the Triangle ZRP.) which ta∣ken out of Angle AZP 130° 03', leaves the Angle AZR 65° 44': Now the Angle A is found by this Analogy or Proportion.

As S. V. PZR, 64° 19',

To S. V. AZR 65 44,

So is Sc. V. at P 31 34,

To Sc. V. at A 30 28: which was required to be found out and known.

PROP. XI. Case 11. Three Sides given, to find an Angle.

In the Triangle APZ, the Side AZ is 40° 00', the Side ZP is 38° 30', the Side AP is 70° 00', and the Angle Z is required. To find which do thus.

Add the three Sides: together, and from half* 1.75 their Sum, deduct the Side opposite, to the re∣quired Angle: and then proceed as you see in the Operation following.* 1.76* 1.77

½Sum is 65° 07' 30", the Sc. ½ V. at Z which doubled is 130° 03' 12"; the Angle at Z required.

PROP. XII. Case 12. Three Angles given, to find a Side.

In the Triangle AZP, the Angle A is 30° 28'* 1.78 11", the Angle Z 130° 03' 12", the Angle P is 31° 34' 26", and the Side AZ, opposite to P, is required.

This Case is likewise performed as the former Case or Proposition, the Angles being conver∣ted into Sides, and the Sides into Angles, by taking the Complement of the greatest Angle unto 180°: see the work.* 1.79* 1.80 which being doubled, gives the Side AZ 40° 00 required to be found out and known

☞ But if the greater Side AP were required the Operation would produce the Complem〈…〉〈…〉 thereof unto a Semicircle or 180°; therfo〈…〉〈…〉

substract it from 180°, it leaves the remaining required Side sought.

Thus I have laid down all the Cases of Tri∣angles, both Right-lined and Spherical; either Right, or Oblique-angled; I might hereunto have annexed many Varieties unto each Case, and some fundamental Axioms, which somewhat more would have Illustrated and Demonstrated those Cases, and Proportions; but because of the smallness of this Treatise, which is intended more for Practice than Theory, I have for brevi∣ty sake omitted them, and refer you for those things to larger Authors, who have largely discoursed thereon to good purpose.

Notes

• * 1.1

  Fig. 34.
• * 1.2

  Fig. 36.
• * 1.3

  Fig. 35.
• * 1.4

  Fig. 34.
• * 1.5

  〈 math 〉〈 math 〉
• * 1.6

  Observe this for a ge∣neral Rule in Trigonome∣try.
• * 1.7

  Fig. 34
• * 1.8

  Fig. 34.
• * 1.9

  Fig. 34.
• * 1.10

  Fig. 35.
• * 1.11

  〈…〉〈…〉
• * 1.12

  Fig. 35.
• * 1.13

  Fig. 35.
• * 1.14

  Fig. 36.
• * 1.15

  ☞ The Rule to find the Complement Arithmeti∣cal, of any Logarithm Number.
• * 1.16

  9. 962398.
• * 1.17

  0. 037602.
• * 1.18

  [〈 math 〉〈 math 〉]
• * 1.19

  Fig. 36.
• * 1.20

  Fig. 36.
• * 1.21

  Fig. 36.
• * 1.22

  Fig. 37.
• * 1.23

  Fig. 37.
• * 1.24

  That is equal unto the two Angles B 40° and B. 53° as afore in the former Proposition.
• * 1.25

  [〈 math 〉〈 math 〉]
• * 1.26

  Fig. 37.
• * 1.27

  〈 math 〉〈 math 〉
• * 1.28

  Fig. 37.
• * 1.29

  Fig. 37.
• * 1.30

  Fig. 37.
• * 1.31

  Fig. 38.
• * 1.32

  〈 math 〉〈 math 〉
• * 1.33

  Fig. 38.
• * 1.34

  Fig. 38.
• * 1.35

  Fig. 38.
• * 1.36

  Fig. 38.
• * 1.37

  Fig. 38.
• * 1.38

  Fig. 38.
• * 1.39

  Fig. 38.
• * 1.40

  Fig. 38.
• * 1.41

  Fig. 38.
• * 1.42

  Fig. 38:
• * 1.43

  Fig. 38.
• * 1.44

  Fig. 38.
• * 1.45

  Fig. 38.
• * 1.46

  Fig. 38.
• * 1.47

  Fig. 38.
• * 1.48

  Fig. 39.
• * 1.49

  Note that in thes•• Operations, for the more facility of the learner, I omit Seconds, which doth belong unto the Angles, &c.
• * 1.50

  Fig. 39.
• * 1.51

  Fig. 39.
• * 1.52

  〈 math 〉〈 math 〉
• * 1.53

  ☞ And here ob∣serve that if the Sum of the two contained Sides exceed a Semicircle, then substract each side severally from 180°, and proceed with those Complements, as with the sides given, the Operation produceth the Complements of the Angles sought, unto a Semicircle or 180 Degrees.
• * 1.54

  Fig. 39.
• * 1.55

  ☞ And here observe that if the Sum of the two given Angles excede a Semicircle or 180°, substract them from a Se∣micircle, and proceed with the Residues, the Operation will produce each side's Complement to a Semicircle, or 180 Degrees.
• * 1.56

  Fig. 39.
• * 1.57

  〈 math 〉〈 math 〉
• * 1.58

  Fig. 39.
• * 1.59

  〈 math 〉〈 math 〉
• * 1.60

  〈 math 〉〈 math 〉
• * 1.61

  Fig. 39.
• * 1.62

  〈 math 〉〈 math 〉
• * 1.63

  〈 math 〉〈 math 〉
• * 1.64

  Fig. 39.
• * 1.65

  〈 math 〉〈 math 〉
• * 1.66

  〈 math 〉〈 math 〉
• * 1.67

  Fig. 39.
• * 1.68

  Fig. 39.
• * 1.69

  〈 math 〉〈 math 〉
• * 1.70

  〈 math 〉〈 math 〉
• * 1.71

  Fig. 40.
• ☞ 1.72

  Note that if the Angles at the Base be both of one kind, that then the Perpendicular falls within the Triangle: if of diverse kinds, without the Triangle.
• * 1.73

  Fig. 41.
• * 1.74

  Fig. 40.
• * 1.75

  Fig. 40.
• * 1.76

  〈 math 〉〈 math 〉
• * 1.77

  Fig. 40.
• * 1.78

  Fig. 40.
• * 1.79

  〈 math 〉〈 math 〉
• * 1.80

  Fig. 40.