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. III. (Book 3)

Of the Construction of the Canon of Triangles. (Book 3)

THat the Proportions which the parts of a Triangle have one to another may be certain, the arches of circles (by which the an∣gles of all Triangles, and of Spherical Triangles the sides are also measured) must be first reduced into right lines, by defining the quantity of right lines, as they are applyed to the arches of a circle.

2 Right lines are applyed to the arches of a circle three wayes, viz. ei∣ther as they are drawn within the circle, without the circle, or as they are drawn through it.

3 Right lines within the circle are Chords and sines.

4 A Chord or subtense is a right line inscribed in a ci••cle, dividing the whole circle into two segments: and in like manner subtending both the

segments: as the right line CK divi∣deth the circle GEDK into the two segments CEGK and CDK, and subtendeth both the segments, that is, the right line CK is the chord of the arch CGK, and also the chord of the arch CDK.

5 A Sine is a right line in a semi∣circle falling perpendicular from the term of an arch.

6 A Sine is either right or versed.

7 A right Sine is a right line in a Semicircle, which from the term of an arch is perpendicular to the diameter, dividing the Semicircle into two seg∣ments, and in like manner referred to both: Thus the right line CA is the sine of the arch CD less then a qua∣drant, and also the sine of the arch CEG greater then a quadrant, and hence instead of the obtuse angle GBC, we take the acute angle CBA the complement thereof to a Semicir∣cle, and so our Canon of Triangles doth never exceed 90 deg.

8 A right sine is either Sinus totus, that is, the Radius or whole Sine, as the right line EB: or Sinus simplici∣ter the first sine, or a sine less then Radius, as AC or AB, the one where∣of is alwayes the complement of the other to 90 degrees, we usually call them sine and co-sine.

9 A Versed sine is a right line in a Semi-circle perpendicular to the right

[illustration]

Sine of the same arch; Thus A D is the versed sine of the arch CD and GBA is the versed sine of the arch

GEC the complement of CD.

10 Right lines without the Circle whose quantity we are to define, are such as touch the circle, and are cal∣led Tangents.

11 A Tangent is a right line which touching the circle without, is perpen∣dicular from the end of the diameter to the Radius, continued through the term of that arch of which it is the Tangent: Thus the right line FD is the Tangent of the arch CD.

12 Right lines drawn through the circle, whose quantity we are to define are such as cut the circle and are cal∣led Secants.

13 The Secant of an arch, is a right line drawn through the term of an arch, to the Tangent line of the same arch: and thus the right line BF is the Secant of the arch CD: as also of the arch CEG the complement thereof to a Semicircle.

14 A Canon of Triangles then is that which conteineth the Sines, Tan∣gents, and Secants of all degrees &

parts of degrees in a quadrant, accor∣ding to a certain diameter, or measure of a circle assumed: The constructi∣on whereof followeth, and first of the Sines.

15 The right Sines as they are to be considered in order to their con∣struction are either Primary or Se∣condary.

16 The Primary Sines are those by which the rest are found: And thus the Radius or whole sine is the first pri∣mary sine, and is equal to the side of a six-angled figure inscribed in a Circle.

Demonst. AC = AB by the work,

[illustration]

and ang. C = B by the second of the second hereof, and BC the measure of the angle at A is 60 degr. and the angles A, B, C, are together equal to two right angles, and therefore ang. A = C or B and BC = AC or AB as was to be proved.

Consectary.

The Radius of a circle being given, the sine of 30 deg. is also given, for by this proposition, the Radius of a circle is the subtense of 60 deg. and the half thereof is the sine of 30, and therefore the Radius AB, or BC be∣ing 1000.0000 the sine of 30 deg. is 500.0000.

17 The other primary sines are the sines of 60.18 and 12 deg. being the half of the subtenses of 120. 36 & 24 degr. and may be found by the problems following.

18 The right sine of an arch & the right sine of its complement, are in power equal to Radius.

Demonst. In the first diagram of this chapter, AC is the sine of CD,

and AB the sine of CE the comple∣ment thereof, which with the Radius BC make the right-angled Triangle ABC, therefore ABq + ACq = BCq by the 19 of the second, as was to be proved.

And hence the sine of 60 deg. may thus be found, let the sine of 30 deg. AC be 500.0000 the square where∣of 250.00000 being subtracted from the square of BC Radius, the remainer is 750.00000 the square of AB, whose square root is 8660254 the sine of 60 deg.

19 The subtense of 36 deg. is the side of a Dec-angle inscribed in a circle, or the greater segment of a Hexagon divided into extream and meane proportion.

Demonst. Let DG = DA and EG

[illustration]

bisect the ang. AGD = DAG by the

second of the second, and the angle DGE = ADG because AGD half the complement of ADG (36 deg.) to a Semicircle is bi-sected by the right line EG = DE by the second of the second, and the angle AEG = EGD + EDG, and therefore AEG = EAG = AGD and EG = AG by the se∣cond of the second, and the Triangles ADG and EAG like, therefore AD, AG = ED ∷ AG AE, and so AD the side of a Hexagon is divided into ex∣tream and meane propo••tion, & ED the greater segment equal to AG the side of a dec-angle, as was to be pro∣ved.

Corsectary.

The side of a Hexagon being given, the side of a Dec-angle or subtense of 36 degr. is also given, for by the se∣cond of the second the Semi-radius being deducted from the square root of the squares of Radius, and the half Radius added together, the remainer is 6180339 the subtense of 36 deg.

20 The subtense of 24 deg. is the side of a Quin-decangle, or the difference between the subtenses of 60 and 36 deg. and may be found by the 24 of the second.

Demonst. Let AC be the subtense of 60 deg. AB 36, then is BC 24 the difference, BD the complement of AB and CD the complement of AC

[illustration]

and AD the diameter. And AC × BD = AB × CD + BC × AD there∣fore AC × BD − AB × CD = AD × BC, and AD × BC being divided by AD the diameter, the quotient is BC the subtense of 24 deg.

21 Having thus found the primary sines, the secondary sines as the sines of 6 d••g 3 deg. 1 d. 50 cent. 0 deg. 75 centesm••s, and 0 d. 01 cen∣tesme may be found from them, and all the other sines, to every degree, and part of a degree in the quadrant.

22 The right sine and versed sine of an arch are together equal in power to the subtense of the same arch.

Demonst. The right sine CF and the versed sine FE with the subtense EC doe make a right angled Trian∣gle,

[illustration]

and therefore FCq + FEq = ECq by the 19 of the second, as was to be proved,

Consectary.

The right sine and versed sine of an arch being given, the sine of half that arch is also given, for EH the ½ EC is the sine of ED the halfe of the arch EDC.

23 The right sine of an arch is a meane pro∣protional between the Semi-radius, and the versed sine of the double arch.

Demonst. In the preceding diagram the Triangles EHA and ECF are like, because ang. F = EHA and E common to both, therefore

AE, EC ∷ EH, EF or AE. EC ∷ ½ AE : ½ EC that is, ½ AE, ½ EC = EH ∷ EH, EF as was to be proved.

1 Consectary.

Therefore by this or the former proposition, the fine and sine comple∣ment of an arch being given, the sine of half that arch is also given, I say the sine complement, because the ver∣sed sine is found, by deducting it from Radius; Thus FA the sine of BC being deducted from AE Radius the remainer is FE the versed sine of EDC and ½ AE × EF = EHq whose root is EH or the sine of ED.

24 The rectangle of the sine and sine comple∣ment of an arch is equal to the rect••ngle of half the Radius, and the sine of the double arch.

Demonst. In the preceeding dia∣gram, EH is the sine of ED and HA the cosine thereof, and CF is the sine of EDC the double arch, the Trian∣gle HAE and EFC are like, as be∣fo••e.

And AE, EC ∷ HA, CF And ½ AE, EH = ½ EC ∷ HA CE And the rectangle of AH × HE = ½ AE × CF as was to be proved.

25 By these Propositions the sine of 12 deg. being given the sines and sin••s complem••nts of these arches 6 deg. 3 deg. 1 deg. 50 cent. 0 deg. 75 cent. &c. were found to be as follow∣eth.

Deg. parts	Sines	Co-sines.
6.00	10452.846••2	99452.18953
3.00	5233.59562	99862.95347
1.50	2617.691••3	99965.73249
0.75	1308.95955	99991.43••75
0.375	654.49379	99997.85816
0.1875	327.24865	99999.46453

And from the sine of 0 deg. 1875 parts, the sine of 10 centesmes may be found, in this manner.

• As, 0 deg. 1875
• Is to the sine thereof 327.14865
• So is 0 deg. 10 centesmes.
• To the sine thereof 174.53261
• Therefore the sine of 0d. 5 cent. 87.26930
• The sine of 0d. 01 cent. is 17.43326
• And the Co-sine thereof is 99999.99847

Which being given the rest of the Ta∣ble of sines may be ea••••ly made by this proposition following.

26 Three equi-different arches being pro∣pounded, the rectangle made of the cosine of the common difference and the double sine of the meane arch, is equal to the R••ctangle made of the Radius, and the summe of the sines of the two extream arches.

Demonst. Let the three equi-diffe∣rent arches be ED, EC, & EB, whose common difference is BC o•• CD the arch EF = DE, therefore the arch BEF = 2 CE, and B•• = 2 C•• the sine

of CE and GF = BD the double measure of the ang. GBF = BAC and the ang. BNA & BHF right, therefore the Triang. ABN and BFH like AB, AN ∷ BF, BH the sum of BM and BK the sines of the extream arch∣es, therefore

[illustration]

AN × BF = AB × BH which was to be proved.

Consectary.

The sines of ED, and EC with the co-sine of CD or BC being given, the sine of EB is also given, for if from BH the sum of BM and DK you de∣duct MH = DK the remainer is BM the sine of BE, for illustration sake we have added these examples following.

1 Example.

ED 5 cent. and EC 10 centesmes being given to find BM the sine of EB 15 centesmes.

• As the Radius AB 10000.0000
• To AN the cos. BC 5′ viz. 9999.9962
• So BF the doub. of LC 10′ 349.0650
• To BH = BM + DK 349.0659
• Frō which subt. MH = DK 5′ cent. 87.2663
• Resteth the sine of 0d 15 cent. BM 261.7987

2 Example.

ED 0 deg. 10 cent.

EC 0 deg. 15 cent. given, to find EB 0 deg. 20

• As the Radius AB 10000.0000
• To cos. BC 5 cent. viz. AN 9999.9962
• So the double of LC 15 cen. viz BF. 523.5974
• To BH = BM + DK 523.5976
• Subt. MH = DK 10 cent. 174.5326
• Rest sine 0 deg. 20 cent. BM. 349.0650

3 Example.

Let EB 0 d. 25 cent. be inquired. There being given ED 0 deg. 15 cent. EC 0 deg. 20 cent.

• As the Radius AB 90 10000.0000
• To Cos. BC viz. s AN 89d 95 9999••9962
• So the doub. of LCs 0d 20 cent. 698.1300
• To BH = BM + DK 698.1297
• ...Sub. MH = DK 0 d. 15 cent. ••61.798••
• Rest BM the sine of 0d. 25 cent. 436.330••

4 Example.

Let EB 0 d. 30 cent. be inquired. There being given ED 0 d. 20 cent. EC 0 d. 25 cent.

• As the Radius 90 1000.0000
• To Cos. BC viz. s AN 89.95 9999.9962
• So twice L•• = BF viz. 25 cent. 872.6618
• To BH = BM + DK 872.6613
• Subt. DK = MH 0d. 20 cent. 149.0950
• Rest▪ MB the sine of 0 d. 30 cent. 523.5963

And in this manner may the whole Canon of sines be compleated, or the sines of the first or last 60 deg. being thus made, or the sines of 45 deg. be∣in•• made by this rule and the sines of 15 d more by the 24th thereof, the other 30 d. may be more easily made by the following Probleme.

26 The sum of the Sines of any two arches equally distant from 30 deg. is equal to the co∣sine of the distance.

Demonst. In the preceding diagram we have already proved.

AB. AN ∷ BF. BH therefore also ½ AB. AN ∷ ½ BF. BH, now then if ½ BF = ½ AB sine of 30 deg. AN the cosine of the distance must be also equal to BH, the sum of the sines, as was to be proved.

Example.

Let DK the sine of DE 27 d. be 45399.04997

And BM the sine of EB 33 d. be 54463.90350

Their sum is BH = AN sine of 87d. 99862.95347

27 The Canon of sines being thus made, a Table of Tangents and Secants may be easily deduced from them, by the following problemes.

28 As the co-sine of an arch, is to the sine thereof, so is Radius, to the Tangent of that arch.

Demonst. In the annexed diagram, the Triangles AEF and AHG are like, because of their right angles at F and G, & their common angle at A. Therefore, AF. FE ∷ AG. GH.

[illustration]

29 The Radius is a meane proportional, be∣tween the tangent and the tangent complement of an arch.

Dem. HG is the tangent of an arch, CK the co-tang. thereof & LH = AG and the triangles, ALH and ACK are like, because of their right an∣gles at L and C, and their common angle at A. Therefore AL = HG. LH ∷ AC. CK.

30 The Radius is a meane proportional, be∣tween the right sine of an arch, and the secant of its complement.

Demonst. In the preceding diagram the triangles AEF & AHG are like, therefore, AF. AE ∷ AG. AH.

31 As the sine of an arch or angle is to Rad. so is the tangent of the same arch, to the secant thereof.

Demonst. In the preceding diagram

the triangles AEF & AGH are like, therefore EF. AE ∷ HG. AH.

32 As Radius, is to the secant of an arch, so is the co-tangent of the same arch, to the co-secant thereof.

Demonst. In the preceding diagram, the triangles ALH and ACK are like, therefore LH, AH ∷ CK. AK.

Other more easie and expedi∣tious wayes of making the Tangents and Secants, you may see in the first Chap. of my Trigonometria Britan∣nica, but the Canons being now al∣ready made, these Rules we deeme sufficient.

The construction of the Artificial Sines and Tangents, we have purpose∣ly omitted, they being nothing els but the Logarithmes of the Natural, of which Logarithmes we have shew∣ed the construction in a former Insti∣tution, by the extraction of roots, and in my Trigonometria Britannica by multiplication: and therefore shall now proceed to the use of the Canon of Sines, Tangents and Secants, in

the solution of all Triangles, whether plaine or Spherical.