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. IV. (Book 4)

Of the Calculation of plain Triangles. (Book 4)

A Plain Triangle is contained un∣der three right lines, and is ei∣ther Right-angled, or Obli∣que.

2 In all plain Triangles, two an∣gles being given the third is also gi∣ven: and one angle being given the sum of the other two is given: because the three angles together are equal to two right by the 9th of the second.

Therefore in a plain right angled triang. one of the acute angles is the complement of the other.

3 In the resolution of plain Trian∣gles, the angles onely being given, the sides cannot be found, but onely the the reason of the sides: It is therefore necessary, that one of the sides be known.

4 In a Right-angled Triangle two termes (besides the right-angle) will serve to find the third; so the one of them be a side.

5 In Oblique angled Triangles there must be three things given to find a fourth.

6 In right-angled plain Triangles there are seven cases, and five in Ob∣lique, for the solution of which the four Axiomes following are suffi∣cient.

1 Axiome.

In a right-angled plain Triangle: The rectangle made of Radius & one of the sides containing the right-angle, is equal to the rect-angle made of the other containing side, and the tangent of the angle thereunto adjacent.

Dem. In the right angled plain triang. BED draw the periphery FE, then is BE Radius & DE the Tangent of the angle at B, make CA parallel to DE, then are the Triangles ABC and EBD like, because of their right∣angles at A and E and their common

angle at B, therefore BA, BE ∷ AC. ED. and BA × ED = BE × AC. that

[illustration]

is BA × tB = Rad. × AC. as was to be proved.

2 Axiome.

In all plain Triangles: The sides are proportional, to the sines of their opposite angles.

Demonst. In the ••plain triangle BCD extend BC to F making BF = DC, and draw the arches FG & CH, then are the perpendiculars FE & CA the sines of the angles at B & D by the 7th of the third, and the triangles BEF and BAC are like, because of their right angles at E and A, and their common angle at B. Therefore

C. C A ∷ BF. FE. that is

[illustration]

BC. sine D ∷ DC = BF. sine B. as was to be proved.

3 Axiome.

In all plain Triangles; As the halfe sum of the sides, is to their halfe difference: so is the tangent of the half sum of their opposite angles, to the tangent if their half difference.

Demonst. In the triangle BCD let the sides be CB and CD, and CG = CB. wherefore ½ Z crur. = EG. & ½ × crur. = EC, draw CH bi-se∣cting BG at right angles, and make the angle GCI = D, then will the angle GCH = ½ Z angle B and D whose tangent is HG, and the angle

ICH = ½ × ang. B and D whose tan∣gent is HI.

[illustration]

But EG. EC ∷ HG. HI. that is. ½ Z crur. ½ × crur. ∷ t ½ Z ang t ½ X ang.

4 Axiome.

In all plain triangles: As the base, is to the sum of the other sides, so is the difference of those sides, to the difference of the segments of the base.

Demonst. In the triangle BCD let fall the perpendicular AC, extend

BC to F and draw FG and DH, then •• BF = Z crur. and HB = × crur. & GB the difference between AB and AD the segments of DB the base. And the triangles HDB and BGF are

[illustration]

like, because of their equal ang at D & F the arch HG being the double mea∣sure to them both, and their common angle at B. Therefore

DB. BF ∷ HB. GA. That is, DB. Z crur. ∷ × crur. × seg. base.

These things premised, we will now set down the several Problems or ca∣ses

in all plaine triangles right-an∣gled and oblique, with the proporti∣ons by which they may be solved, & manner of solving them both by na∣tural and Artificial numbers.

Of right angled Plain Triangles.

IN right angled plain triangles, the sides comprehending the right-an∣gle we call the Legs, and the side sub∣tending the right angle, we call the Hypotenuse.

1 Probl. The legs given to find an Angle.

In the right-angled plain triangle,

[illustration]

ABC, the angle ACB is inquired.

The given legs

• AB. 230
• AC. 143.72

AC. Rad ∷ AB. tA CB by the 1 Axi∣ome.

That the quantity of this angle, or any other term required may be ex∣pressed in numbers, if the solution be to be made in natural numbers, mul∣tiply the second term given by the third, and their product divide by the first, the Quotient is the fourth pro∣portional sought.

But if the solution be to be made in artificial numbers, from the sum of the Logarithmes of the second and third terms given, subtract the Loga∣rithme of the first, the remainer shall be the Logarithme of the fourth pro∣portional required.

Illustration by natural numbers.

• As the Leg AC 143.72
• Is to the Radius AC 10000000
• So is the Leg. AB 235
• To the tang. of ACB grad. 58.55

The product of the second and third

termes is 2350000000, which being divided by 143.72 the first term given, the quotient is 16351238 the tangent of the angle ACB which being sought in the Table the neerest less is 16350527, and the arch answering thereto is Grad. 58. 55 parts.

Illustration by Logarithmes.

	Logarithmes.
As the leg AC 143.72	2.157517
Is to the rad. IC	10.0000
So is the leg AB 235	2.371068
To the tang. of. C. gr. 58.55	10.213551

2 Probl. The angles and one leg gi∣ven, to find the other leg.

In the right angled plain Triangle ABC, the leg AC is inquired:

The given

• Angle ABC
• Leg. AB

Rad. AB▪ ∷ t ABC. AC. by the first Axiome.

3 Prob. The Hypotenuse and a leg given to find an angle.

In the right angled plain Triangle ABC, the angle ACB is inquired.

The given

• Hypoth. BC.
• Leg AB.

BC. Rad ∷ AB. s. ACB. by 2 Ax.

4 Probl. The Hypotenuse and angles given, to find either leg.

In the right angled plain Triangle ABC, the leg. AB is inquired:

The given

• Hypot. BC.
• Angle ACB.

Rad. BC ∷ s. ACB. AB. by 2 Ax.

5 Probl. The angles and a leg gi∣ven, to find the Hypotenuse.

In the right-angled plain Triangle ABC, the Hypotenuse BC is inquired;

The given

• Angle ABC.
• The given Leg AC.

s. ABC. AC ∷ Rad. BC. by the se∣cond Axiome.

6 Probl. The Hypotenuse and leg given, to find the other leg.

In the right-angled plain Triangle ABC, the leg AB is inquired,

The given

• Hypot. BC
• Leg. AC

1. BC. Rad ∷ AC. s. ABC, by the 3 Problem.

2. t. ABC. AC ∷ Rad. AB. by the 2 Probl.

7 Probl. The legs given, to find the Hypotenuse.

In the right-angled plain Triangle ABC the Hypotenuse BC is inquired.

The given legs

• AB
• AC

1 AB. Rad ∷ AC. t. ABC by the 1 Problem.

2 s. ABC. AC ∷ Rad. BC. by the 5 Problem.

Of Oblique angled plaine Triangles.

1 Probl. Two sides and an angle op∣posite to one of them given, to find the an∣gle opposite to the other side.

In the Oblique angled plain Trian∣gle DCB the angle CBD is inquired.

The given Sides

• DC 865
• CB 632

The given Angle CDB 26. 37

[illustration]

But here it must be known whether the angle sought, be acute or obtuse. CB. s CDB ∷ CD, s. CBD. by the second Axiome.

Illustration by natural Numbers.

• As the side CB 632
• Is to the sine of CDB 26.37 4441661
• So is the side CD 865
• To the sine of CBD 37.43 6079172

For if you multiply 4441661 by 865, the product will be 3842036765, which product being divided by the first term 632, the quotient 6079172 is the sine of 37 deg. 43 parts.

Illustration by Artificial numbers.

• As the side CB 632 2.800717
• Is to the sine of D 26.37 9.647545
• So is the sides CD 865 2.937016
• 12.584561
• To the sine of B 37.43 9.783844

2 Probl. Two sides with the angle comprehended by them given, to find ei∣ther of the other angles.

In the Oblique angled plain Tri∣angle BDC.

The angle DBC is inquired.

The given Sides

• DC
• BC

The given Angle DCB

Subtract the angle given, out of 180d. the remainer is the sum of the other angles, and ½ Z c••ur. DC and BC. ½ X crur ∷ t ½ Z ang. t ½ X ang. by the 3 Ax. To the half sum of the other angles, adde the half difference found, and you will have the greater angle, sub∣duct it & you will have the lesser, in our example ½ Z ang. + ½ Xang. is = DBC sought, because that is the greater ang.

3 Probl. The angles and a side gi∣ven, to find either of the other sides.

In the Oblique angled plain triangle CBD,

The side DB is inquired.

The given Angles

• DCB
• CDB

The given Side CB

s. CDB, CB ∷ s. DCB. DB. by the second Axiome.

4 Probl. Two sides with the angle comprehended by them given, to find the third side.

In the oblique angled plain triangle DCB

The side CB is inquired.

The given Sides

• DB
• CB

The given Angle CBD

First, find the angle BCD by the third Axiome.

Secondly, find the side CB by the se∣cond Axiome.

5 Probl. The three sides given to find an angle.

In the Oblique angled plain trian∣gle CBD.

The angle CDB is inquired.

The three sides be∣ing given.

• DC
• CB
• DB

The resolution of this Problem doth require two operations, first, for the segment of the base GB.

DB.Z crur. DC & CB ∷ X crur. GB. ••y the fourth axiome.

And DB − GB = DG, and ½ DG =DA or AG.

And DC. Rad ∷ AD. s ACD, by the second axiome whose complement ••s ADC.

And AG + GB = AB, therefore ang. ABC may be also found in like manner.