Geometrical trigonometry, or, The explanation of such geometrical problems as are most useful & necessary, either for the construction of the canons of triangles, or for the solution of them together with the proportions themselves suteable unto every case both in plain and spherical triangles ... / by J. Newton ...

CHAP. VI. (Book 6)

Of the Calculation of Spherical Triangles. (Book 6)

IN Spherical Triangles there are 28 varieties or cases, 16 in rectangular and 12 in oblique angular, whereof all the rectangular, and 10 of the ob∣lique angular may be resolved by these two Axiomes following.

1 Axiome.

In all Spherical rectangled Trian∣gles, having the same acute angle at the base: The sines of the Hypotenu∣saes are proportional to the sines of their perpendicular.

2 Axiome.

In all Spherical rectangled Trian∣gles, having the same acute angle at the base: The sines of the bases, and the tangents of the perpendiculars are proportional.

Demonst. Let ADB and AIM represent two Spherical Triangles ha∣ving

the same angle at A, then is IH the sine of IM and FD the sine of DB. But

s. AD. AI Rad ∷ s DF s. IH.

[illustration]

Again. KB is the tang. of DB & LM the tang. of IM. And

s. AB. t. KB ∷ s. AM. t. LM. as was to be proved.

That all the case of a right-angled Spherical Triangle may be resolved, by these two Axiomes, the several parts of the Spherical triangle, pro∣posed must sometimes be continued to quadrants, that so the angles may

be turned into sides, the hypotenusaes into bases and perpendiculars, and the contrary, by which meanes, the pro∣portions as to the parts of the Trian∣gle given, instead of sines doe some∣times fall in Co-sines, and sometimes in co-tangents instead of tangents, which the L. Nepeir observing, those parts of the right-ang. Spherical trian∣gle which in such conversion doe for the most part change their propor∣tion, he noteth with their comple∣ments, viz. the hypotenuse and both the acute angles, but the sides con∣taining the right angle are not so no∣ted: and these five he calleth the Cir∣cular parts of the triangle, amongst which the right angled is not reckon∣ed, and therefore the two sides which doe contain it, are supposed to be joyned together.

Of these five Circular parts, one is alwayes in the middle, and two of the five are adjacent to that middle part, and the other two are disjunct, the parts adjacent, are called extreams

adjacent, and the parts disjoyned, are called extreames disjunct.

Each of these five circular parts, may by supposition be made the midle part, and then the two circular parts which are next to that which is by supposition made the middle, are the extreames conjunct, the other two remote from the middle part assumed are the extream parts disjoyned.

As in the Triangle ABC, if comp. AC be made the middle part, comp. A, and comp. C are the extreames conjunct, and the sides AB and BC are the extreames disjunct, and so of the rest as in the Table following.

[illustration]

Mid. part	Extr. conj.	Extr. disj.
Leg. AB.	Comp. A Leg BC	Comp. AC Com. C
Comp. A	Comp. AC Leg. AB	Com. C Leg. BC
Comp. AC	Comp. A Comp. C	Leg. AB Leg. BC
Comp. C	Comp. AC Leg. BC	Comp. AB Leg. AB
Leg. BC	Comp. C Leg. AB	Comp. A Comp. AC

The parts of a right angled Sphe∣rical Triangle, being thus distinguish∣ed into five circular parts, (besides the right angle) the three parts remote from the right angle, being noted by their complements, and of these five one accounted the middle, the other either extreams adjacent or disjunct, the L. Nepeir as a consectary from the two preceding axiomes, for the help of memory, and therefore the more easie resolving of all spherical Triangles

hath composed this Catholick and Universal Proposition.

The Sine of the Middle part and Radius are reciprocally proportional, with the tangents of the extreams con∣junct, and the co-sines of the extreames disjunct.

That is, As Radius, to the tangent of one of the extreames conjoyned: So is the tangent of the other extream conjunct to the Sine of the Middle part.

And also, As Radius, to the co-sine of one of the extreames disjunct: So is the co-sine of the other extream disjunct, to the sine of the Middle part.

Therefore, if the Middle part be sought, the radius must be in the first place, If either of the extreams, the o∣ther extream must be in the first place.

Onely note, That if the Middle part, or either of the extreams conjunct, be noted with its complement in the cir∣cular parts of the Triangle, instead of the sine or tangent, you must use the co-sine or co-tangent of such circu∣lar part or parts.

If either of the extreams disjunct, be noted by its complement in the cir∣cular parts of the triangle, instead of the co-sine, you must use the sine of such extream disjunct.

That these directions may be the better conceived, we have in the table following set down the circular parts of a Triangle under their respective titles, whether they be taken for the Middle part, or for the extreams, whe∣ther conjunct or disjunct, & unto those parts we have perfixed the sine or co∣sine, the tangent or co-tangent as it ought to be by the former Rules.

Middle part	Extr. conjunct	Extr. disjunct.
Sine AB	Co-tang A	Sine AC
Tang. BC	Sine C
Cosine A	Co-tang. AC	Sine C
Tang. AB	Co-sine BC
Co-sine AC	Co-tang. A	Co-sine AB
Co-tang. C	Co-sine BC
Co-sine C	Co-tang. AC	Sine A
Tang. BC	Co-sine AB
Sine BC	Co-tang. C	Sine A
Tang. AB	Sine AC

These things premised, we will now set down the several cases with the Analogies by which they may be sol∣ved, according to this Catholick pro∣position of right angled Spherical tri∣angles first, and then of oblique, and in every case in which we deeme it ne∣cessary, we will Demonstrate the truth of the Proposition by the first & second Axiomes of this chapter, as where the middle part, is either one of the legs containing the right angle, or one of the oblique angles or the Hypote∣nusal.

Of right angled Spherical Triangles, Problem. 1.

A leg with an angle opposite thereunto being given, to find the other leg. If it be known whe∣ther the Hypoteuuse or the other angle be more or less then a quadrant.

In the right angled Spherical Tri∣angle ABC.

The leg AB or middle part is inquired

The given

• Leg BC
• ang. com. A

are extream conjunct.

Analogie Rad. cot. ang. A ∷ tang. BC. sine AB.

Demonst. The Trian∣gles AGD and ACB have the same acute angle at the base.

[illustration]

Therefore ••an. DG. AD ∷ tang. BC ••ine AB by the second Ax. And by the 29th of the third.

Rad. AD. co-tang. DG tang. DG. AD — therefore, Rad. cot. DG ∷ tang. BC. sine AB and DG is the measure of the angle, at A therefore, as Rad. &c.

Illustration by Natural numbers.

• As the Radius. 10000000
• To Co-tang. A grad. 30. 17320508
• So tang. BC. grand. 22.89 4222108
• To sine AB. 46 d. 994 7312904

Illustration by Artificial numbers.

• As the Radius 90 d. 10.000000
• To co-tang. A 30 10.238561
• So tang. BC 22.89 9.625529
• To sine AB. 46.994 9.864090

Problem 2.

A leg and an angle coonterminate therewith being given, to find the other leg

In the R. angled spherical trian. ABC.

The leg BC one of the extreames conjunct is inquired.

The given Angl. Comp. A the other extr. conj. Leg AB the middle part.

Cot. A. Rad ∷ sine AB. tang. BC by the analogie in the first Problem.

Problem 3.

The legs given to find an angle.

In the right angled Spherical Tri∣angle ABC.

The angle comp. A one of the ex∣treams conjunct is inquired.

The given legs

• AB the middle part.
• BC the other extrea. conjunct.

Analogie Tang. BC. sine AB ∷ Rad. cot. A.

The inverse of that analogie in the ••st Probleme.

Problem 4.

The Hypotenuse and a leg given, to find the ••••gle contained by them.

In the right angled Spherical Tri∣••ngle, ABC.

The angle, comp. C, or middle part ••s inquired.

The given

• Hypot. comp. AC
• Leg BC

the adjac. extr.

Analogie, rad. cotang. AC ∷ tang. BC. cosine C.

Demonst. The triangles EFH and EIG have the same acute angle at the base, therefore, ••IG = AC. Rad. IE ∷ t. FH = BC. — sine FE. by the 2 Ax.

Rad. co-tang. IG ∷ tang. IG. Rad. by the 29 of the third. Therefore

Rad. cot. IG ∷ t. FH sine FE, or Rad. cot. AC ∷ t. BC. co-sine C, because FI the complement of FE is the measure of ACB.

Problem 5.

A leg and an angle conterminate with it gi∣ven to find the Hypotenuse.

In the right angled Spherical Tri∣angle ABC.

The Hypot. comp. AC. one of the extreames conjunct is inquired.

The given Leg BC. the other extr. conjunct. Angle comp. C, the middle part.

Analogie. tang. BC. cosine C ∷ rad. cot. AC, it being the inverse of the Analogie in the former Problem.

Problem 6

The Hypotenuse and angle given, to find the leg conterminate with the given angle.

In the right angled Spherical Tri∣angle, ABC.

The leg BC one of the extreames adjacent is inquired.

The given Hypot. comp. AC the other extr. conj. Angle comp. C. the middle part.

Analogie, cotang. AC. Rad ∷ cos. C. tang. BC.

This also is the inverse of the 3 Probleme, and therefore needs no further demonstration.

Problem 7.

The Oblique angles given, to find the Hy∣potenuse.

In the right angled Spherical Tri∣angle ABC.

The Hypot. comp. AC, the middle part is inquired.

The given ang.

• Comp. A.
• Comp. C

The extreams conjunct.

Analogie, Rad cotang. C ∷ cotang. A cosine AC:

Demonst. The triangles CGH & CIF have the same acute angle at the base, CG is the complement of AC, HG the complement of GD the measure of A, and FI the measure of C, but

Tang. FI. CI ∷ tang. GH, sine GC by 2 axiom. therefore,

Tang. C. Rad. ∷ cot. A. cos. AC. And as Rad. cot. C ∷ tang. C. Rad. Therefore Rad. cot. C ∷ cot. A. cosine AC.

Problem 8.

The Hypotenuse and an angle given, to find the other angle.

In the right angled Spherical Tri∣gle ABC.

The angle comp. C, one of the ex∣treams conjunct is inquired.

The given Hypot. comp. AC, the middle part. Ang, comp. A the other extr. conj. Cotang. A. cosine AC ∷ Rad. cot. C, by the Analogie in the preceding Probl.

Problem 9.

The Hypotenuse, and an oblique angle gi∣ven, to find the leg opposite to the given angle.

In the right angled Spherical Tri∣angle ABC.

The leg BC the middle part is in∣quired.

The given Hyp. comp. AC Angle comp. A extream disjunct.

Analogie, Rad. sine A ∷ sine AC sine BC.

Demonst. The Triangles ACB & AGD have the same acute angle at the base, therefore by the first axiome, AG. s. GD ∷ sine AC. sine BC.

Problem 10.

A leg and an angle opposite thereunto, being given, to find the Hypotenuse. If it be known whether it, or the other leg or angle be acute or obtuse.

In the right angled Spherical Tri∣angle ABC.

The Hypotenuse comp. AC one of the extreams disjunct is inquired.

The given Ang. com. A the other extr. d. Leg. BC the middle part.

Anal. Sine A. R. ∷ sine BC. sine C, by the analogie in the preceding Probl.

Problem 11.

The Hypotenuse and a leg given, to find the angle opposite to the given leg.

In the right angled Spherical Tri∣angle ABC.

The angle comp. A one of the ex∣treams disjunct is inquired.

The given Hyp. comp. AC, the other extr. disj. Leg. BC the middle part.

Anal. Sine AC, sine BC ∷ R. sine A by the analagie in the 10 Probl.

Problem 12.

An angle and leg conterminate with it gi∣ven, to find the other angle.

In the right angled Spherical Tri∣angle ABC.

The angle comp. A the middle part is inquired.

• The given Angle comp. C
• The given Leg. BC

Extreames disjunct.

Anal. Rad. sine C ∷ cs BC. cos. A.

Demonst. The triangles CFI and CHG have the same acute angle at the base, therefore by the first axiom•• Rad. FC. sine FI ∷ sine HC. sine HG, and the compl. of HC is the leg BC, & GD the measure of the angle at A is the complement of HG.

Therefore,

R. sine C ∷ co-sine BC. co-sine A.

Problem 13.

An angle and leg opposite thereunto being given, to find the other angle. If it be known whether it, the Hypotenuse or the other leg be acute or obtuse.

In the right angled Spherical Tri∣angle ABC.

The ang. comp. C one of the ex∣treams disjunct is inquired.

The given Leg BC the other extream disjunct. Angle, comp. A the middle part.

Anal. cs BC. cs A ∷ Rad. sine C. by the analogie in the preceding Pro∣bleme.

Problem 14.

The oblique angles given, to find either leg.

In the right angled Spherical Tri∣angle ABC.

The leg BC one of the extreams disjunct is inquired.

The given Angle, comp. C the other extr. disj. Angle, comp. A, the middle part.

Anal. Sine C Rad ∷ cs A. cs BC. the inverse of that analogie in the last Problem.

Problem 15.

The legs given, to find the Hypotenuse.

In the right angled Spherical Tri••∣gle ABC.

The Hypotenuse comp. AC the middle part is inquired.

The given legs

• AB
• BC

Extream disjunct.

Anal. Rad. cs AB ∷ cs BC cs AC.

Demonst. Tsie triangles HBD and HCG have the same acute angle at the base, therefore

Rad. HB. sine BD ∷ sine CH. s CG, But BA is the complement of BC, and BC the complement of CH, and CG the complement of AC.

Therefore, Rad. cs AB ∷ cs BC. cs AC.

Problem 16.

The Hypotenuse and a leg given, to find the other leg.

In the right angled Spherical Tri∣angle ABC.

The leg AB one of the extreames disjunct is inquired.

The given Leg BC the other extream dijunct. Hypot. comp. AC the middle part.

Anal. cs BC. cs AC ∷ R. cs AB, by the analogie in the 15th Probleme.