TRIGONOMETRY

, the art of measuring the sides and angles of triangles, either plane or spherical, from whence it is accordingly called either Plane Trigonometry, or Spherical Trigonometry.

Every triangle has 6 parts, 3 sides, and 3 angles; and it is necessary that three of these parts be given, to find the other three. In spherical Trigonometry, the three parts that are given, may be of any kind, either all sides, or all angles, or part the one and part the other. But in plane Trigonometry, it is necessary that one of the three parts at least be a side, since from three angles can only be found the proportions of the sides, but not the real quantities of them.

Trigonometry is an art of the greatest use in the mathematical sciences, especially in astronomy, navigation, surveying, dialling, geography, &c, &c. By it, we come to know the magnitude of the earth, the planets and stars, their distances, motions, eclipses, and almost all other useful arts and sciences. Accordingly we find this art has been cultivated from the earliest ages of mathematical knowledge.

Trigonometry, or the resolution of triangles, is founded on the mutual proportions which subsist between the sides and angles of triangles; which proportions are known by finding the relations between the radius of a circle and certain other lines drawn in and about the circle, called chords, sines, tangents, and secants. The ancients Menelaus, Hipparchus, Ptolomy, &c, performed their Trigonometry, by means of the chords. As to the sines, and the common theorems relating to them, they were introduced into Trigonometry by the Moors or Arabians, from whom this art passed into Europe, with several other branches of science. The Europeans have introduced, since the 15th century, the tangents and secants, with the theorems relating to them. See the history and improvements at large, in the Introduction to my Mathematical Tables.

The proportion of the sines, tangents, &c, to their radius, is sometimes expressed in common or natural numbers, which constitute what we call the tables of natural sines, tangents, and secants. Sometimes it is expressed in logarithms, being the logarithms of the said natural sines, tangents, &c; and these constitute the table of artificial sines, &c. Lastly, sometimes the proportion is not expressed in numbers; but the several sines, tangents, &c, are actually laid down upon lines of scales; whence the line of sines, of tangents, &c. See Scale.

In Trigonometry, as angles are measured by arcs of a circle described about the angular point, so the whole circumference of the circle is divided into a great number of parts, as 360 degrees, and each degree into 60 minutes, and each minute into 60 seconds, &c; and then any angle is said to consist of so many degrees, minutes and seconds, as are contained in the arc that measures the angle, or that is intercepted between the legs or sides of the angle.

Now the sine, tangent, and secant, &c, of every degree and minute, &c, of a quadrant, are calculated to the radius 1, and ranged in tables for use; as also the logarithms of the same; forming the triangular canon. And these numbers, so arranged in tables, form every species of right-angled triangles, so that no such triangle can be proposed, but one similar to it may be there found, by comparison with which, the proposed one may be computed by analogy or proportion.

As to the scales of chords, sines, tangents, &c, usually placed on instruments, the method of constructing them is exhibited in the scheme annexed to the article Scale; which, having the names added to each, needs no farther explanation.

There are usually three methods of resolving triangles, or the cases of Trigonometry; viz, geometrical construction, arithmetical computation, and instrumental operation. In the 1st method, the triangle in question is constructed by drawing and laying down the several parts of their magnitudes given, viz, the sides from a scale of equal parts, and the angles from a scale of chords, or other instrument; then the unknown parts are measured by the same scales, and so they become known.

In the 2d method, having stated the terms of the proportion according to rule, which terms consist partly of the numbers of the given sides, and partly of the sines, &c, of angles taken from the tables, the proportion is then resolved like all other proportions, in which a 4th term is to be found from three given terms, by multiplying the 2d and 3d together, and dividing the product by the first. Or, in working with the logarithms, adding the log. of the 2d and 3d terms together, and from the sum subtracting the log. of the 1st term, then the number answering to the remainder is the 4th term sought.

To work a case instrumentally, as suppose by the log. lines on one side of the two-foot scales: Extend the compasses from the 1st term to the 2d, or 3d, which happens to be of the same kind with it; then that extent will reach from the other term to the 4th. In this operation, for the sides of triangles, is used the line of numbers (marked Num.); and for the angles, the line of sines or tangents (marked sin. and tan.) according as the proportion respects fines or tangents.

In every case of triangles, as has been hinted before, | there must be three parts, one at least of which must be a side. And then the different circumstances, as to the three parts that may be given, admit of three cases or varieties only; viz,

1st. When two of the three parts given, are a side and its opposite angle.—2d, When there are given two sides and their contained angle.—3d, And thirdly, when the three sides are given.

To each of these cases there is a particular rule, or proportion, adapted, for resolving it by.

1st. The Rule for the 1st Case, or that in which, of the three parts that are given, an angle and its opposite side are two of them, is this, viz, That the sides are proportional to the sines of their opposite angles,

That is,

As one side given	:
To the sine of its opposite angle	::
So is another side given	:
To the sine of its opposite angle.

Or,

As the sine of an angle given	:
To its opposite side	::
So is the sine of another angle given	:
To its opposite side.

So that, to find an angle, we must begin the proportion with a given side that is opposite to a given angle; and to find a side, we must begin with an angle opposite to a given side.

Ex. Suppose, in the triangle ABC, there be given AB = 365 feet, AC = 154.33 f. [angle]C = 98° 3′ to find the other side, and the angles.

Draw AC = 154.33 from a scale of equal parts: Make the angle C = 98° 3′, producing CB indefinitely: With centre A, and radius 365 feet, cross CB in B: Then join AB, and the figure is constructed. Then, by measuring the unknown angles and side, the former by the line of cords or otherwise, and the side by the line of equal parts, they will be found, as near as they can be measured, as below, viz, BC = 310; the [angle]A = 57°1/4; and [angle]B = 24°3/4.

As AB	=	365	log. 2.5622929
To AC	=	154.33	2.1884504
So sin. [angle]C	=	98° 3′	or 81° 57′ 9.9950993
To sin. [angle] B	=	24° 45′	9.6218568
the sum		122 48
taken from		180 00
leaves [angle]A		57 12

Then, again,

As sin. [angle]C	=	98° 3′	log. 9.9956993
To AB	=	365	2.5622929
So sin. [angle]A	=	57° 12′	9.9245721
To BC	=	309.86	2.4911657

In the first proportion, Extend the compasses from 365 to 154 1/3 on the line of numbers; and that extent will reach, upon the line of sines, from 82° to 24 3/4, which gives the angle B. And, in the second proportion, Extend from 98° to 57 1/4 on the sines; and that extent will reach, upon the numbers, from 365 to 310, or the side BC nearly.

2d Case, when there are Given two Sides and their contained angle, to find the rest, the rule is this: As the sum of the two given sides: Is to the difference of the sides:: So is the tang. of half the sum of the two opposite angles, or cotangent of half the given angle: To tang. of half the diff. of those angles.

Then the half diff. added to the half sum, gives the greater of the two unknown angles; and subtracted, leaves the less of the two angles.

Hence, the angles being now all known, the remaining 3d side will be found by the former case.

Ex. Suppose, in the triangle ABC, there be given

the side	AC = 154.33
the side	BC = 309.86
the included	[angle]C =  98° 3′

to find the other side and the angles.

1. Geometrically.—Draw two indefinite lines making the angle C = 98° 3′: upon these lines set off CA = 154 1/3, and CB = 310: Join the points A and B, and the figure is made. Then, by measurement, as before, we find the [angle]A = 57 1/4; [angle]B 24 3/4; and side AB = 365.

As CB + CA  =	464.19	log. 2.6666958
To CB - CA  =	155.53	2.1918142
So tan. 1/2 A + 1/2 B =	40° 58 1/2′	9.9387803
To tan. 1/2 A - 1/2 B =	16  13 1/2	9.4638987
sum gives [angle]A	57  12
diff. gives [angle]B	24  45

Then,

As sin. [angle]B =	24° 45′	log. 9.6218612
To side AC =	154.33	2.1884504
So sin. [angle]C =	98° 3′,	or 81° 57 9.9956993
To side AB =	365	2.5622885

3. Instrumentally.—Extend the compasses from 464 to 155 1/2 upon the line of numbers; then that extent will reach, upon the line of tangents, from 41° to 16° 1/4. Then, in the 2d proportion, extend the compasses from 24° 3/4 to 82° on the sines; and that extent | will reach, upon the numbers, from 154 1/3 to 365, which is the third side.

3d Case, is when the three sides are given, to find the three angles; and the method of resolving this case is, to let a perpendicular fall from the greatest angle, upon the opposite side or base, dividing it into two segments, and the whole triangle into two smaller rightangled triangles: then it will be,

As the base, or sum of the two segments	:
Is to the sum of the other two sides	::
So is the difference of those sides	:
To the difference of the segments of the base.

Then half this difference of the two segments added to the half sum, or half the base, gives the greater segments, and subtracted, gives the less. Hence, in each of the two right-angled triangles, there are given the hypotenuse, and the base, besides the right angle, to find the other angles by the 1st case.

Ex. In the trangle ABC, suppose there are given the three sides, to sind the three angles, viz,

AB = 365	} to find the angles.
AC = 154.33
BC = 309.86

1. Geometrically.—Draw the base AB = 365: with the radius 154 1/3 and centre A describe an arc; and with the radius 310 and centre B describe another arc, cutting the former in C; then join AC and BC, and the triangle is constructed. And by measuring the angles, they are found, viz. [angle]A = 57° 1/4; [angle]B = 24° 3/4; [angle]C = 98° nearly.

2. Arithmetically.—Having let fall the perpendicular CP, dividing the base into the two segments AP, PB, and the given triangle ABC into the two rightangled triangles ACP, BCP. Then,

As AB      =	365	log. 2.5622929
To CB + CA =	464.19	2.6666958
So CB - CA =	155.53	2.1918142
To BP - PA =	197.80	2.2962171
its half =	98.90
1/2 AB =	182.50
sum BP =	281.40
dif. AP =	83.60

Then, in the triangle APC, right-angled at P,

As AC	=	154.33	log. 2.1884504
To sin. [angle]P	=	90°	10.0000000
So AP	=	83.6	1.9222063
To sin. [angle]ACP	=	32° 48′	9.7337559
its comp. [angle]A	=	57° 12

And in the triangle BPC, right-angled at P,

As BC	=	309.86	log. 2.4911655
To sin. [angle]P	=	90°	10.0000000
So BP	=	281.4	2.4493241
To sin. [angle]BCP	=	65° 15′	9.9581586
its comp. [angle]B	=	24  45
Also to [angle]ACP	=	32  48′
add [angle]BCP	=	65  15
makes [angle]ACB	=	98   3

3. Instrumentally.—In the 1st proportion, Extend the compasses from 365 to 464 on the line of numbers, and that extent will reach, on the same line, from 155 1/2 to 197.8 nearly.—In the 2d proportion, Extend the compasses from 154 1/3 to 83.6 on the line of numbers, and that extent will reach, on the sines, from 90° to 32° 3/4 nearly.—In the 3d proportion, Extend the compasses from 310 to 281 1/2 on the line of numbers; then that extent will reach, on the sines, from 90° to 65° 1/4.

The foregoing three cases include all the varieties of plane triangles that can happen, both of right and oblique-angled triangles. But beside these, there are some other theorems that are useful upon many occasions, or suited to some particular forms of triangles, which are often more expeditious in use than the foregoing general ones; one of which, for right-angled triangles, as the case for which it serves so often occurs, may be here inserted, and is as follows.

Case 4. When, in a right-angled triangle, there are given the angles and one leg, to find the other leg, or the hypothenuse. Then it will,

As radius	:
To given leg AB	::
So tang. adjacent [angle]A	:
To the opp. leg BC, and	::
So sec. of same [angle]A	:
To hypot. AC	.

Ex. In the triangle ABC, right-angled at B,

Given the leg AB = 162	}	to find BC
and the [angle]A = 53° 7′ 48″
conseq. [angle] C = 36 52 12	and AC.

1. Geometrically.—Draw the leg AB = 162: Erect the indefinite perpendicular BC: Make the angle A = 53° 1/8, and the side AC will cut BC in C, and form the triangle ABC. Then, by measuring, there will be found AC = 270, and BC = 216.

2. Arithmetically.

As radius	= 10	log. 10.0000000
To AB	= 162	2.2095150
So tan. [angle]A	= 53° 7′ 48″	10.1249372
To BC	= 216	2.3344522
So sec. [angle]A	= 53° 7′ 48″	10.2218477
To AC	= 270	2.4313627

|

3. Instrumentally.—Extend the compasses from 45° at the end of the tangents (the radius) to the tangent of 53° 1/8; then that extent will reach, on the line of numbers, from 162 to 216, for BC. Again, extend the compasses from 36° 52′ to 90 on the sines; then that extent will reach, on the line of numbers, from 162 to 270 for AC.

Note, another method, by making every side radius, is often added by the authors on Trigonometry, which is thus: The given right-angled triangle being ABC, make first the hypotenuse AC radius, that is, with the extent of AC as a radius, and each of the centres A and C, describe arcs CD and AE; then it is evident that each leg will represent the sine of its opposite angle, viz, the leg BC the sine of the arc CD or of the angle A, and the leg AB the sine of the arc AE or of the angle C. Again, making either leg radius, the other leg will represent the tangent of its opposite angle, and the hypotenuse the secant of the same angle; thus, with radius AB and centre A describing the arc BF, BC represents the tangent of that arc, or of the angle A, and the hypotenuse AC the secant of the same; or with the radius BC and centre C describing the arc BG, the other leg AB is the tangent of that arc BG, or of the angle C, and the hypotenuse CA the secant of the same.

And then the general rule for all these cases is this, viz, that the sides bear to each other the same proportions as the parts or things which they represent. And this is called making every side radius.

Spherical Trigonometry, is the resolution and calculation of the sides and angles of spherical triangles, which are made by three intersecting arcs of great circles on a sphere. Here, any three of the six parts being given, even the three angles, the rest can be found; and the sides are measured or estimated by degrees, minutes, and seconds, as well as the angles.

Spherical Trigonometry is divided into right-angled and oblique-angled, or the resolution of right and oblique-angled spherical triangles. When the spherical triangle has a right angle, it is called a right-angled triangle, as well as in plane triangles; and when a triangle has one of its sides equal to a quadrant of a circle, it is called a quadrantal triangle.

For the resolution of spherical Triangles, there are various theorems and proportions, which are similar to those in plane Trigonometry, by substituting the sines of sides instead of the sides themselves, when the proportion respects sines; or tangents of the sides for the sides, when the proportion respects tangents, &c; some of the principal of which theorems are as follow:

Theor. 1. In any spherical triangle, the sines of the sides are proportional to the sines of their opposite angles.

Theor 2. In any right-angled triangle,
As radius	:
To sine of one side	::
So tang. of the adjacent angle	:
To tang. of the opposite side.
Theor. 3. If a perpendicular be let fall from any angl
upon the base or opposite side of a spherical triangle
it will be,
As the sine of the sum of the two sides	:
To the sine of their difference	::
So cotan. 1/2 sum angles at the vertex	:
To tang. of half their difference.
Theor. 4.
As tang. half sum of the sides	:
To tang. half their difference	::
So tang. 1/2 sum [angle]s at the base	:
To tang. half their difference.
Theor. 5.
As cotan. 1/2 sum of [angle]s at the base	:
To tang. half their difference	::
So tang. 1/2 sum of [angle]s at the vertex	:
To tang. half their difference.
Theor. 6.
As tang. 1/2 sum segments of base	:
To tang. half sum of the sides	::
So tang. half difference of the sides	:
To tang. 1/2 diff. segments of base.
Theor. 7.
As sin. sum of [angle]s at the base	:
To sine of their difference	::
So tang. 1/2 sum segments of base	:
To tang. of half their difference.
Theor. 8.
As sin. sum of segments of base	:
To sine of their difference	::
So sin. sum of angles at the vertex	:
To sine of their difference.
Theor. 9.
As sine of the base	:
To sine of the vertical angle	::
So sin. of diff. segments of the base	:
To sin. diff. [angle]s at vertex, when the perp.
falls within	::
Or so sin. sum segments of base	:
To sin. sum vertical [angle]s, where the perp.
falls without.
Theor. 10.
As cosin. half sum of the two sides	:
To cosine of half their difference	::
So cotang. of half the included angle	:
To tang. half sum of opposite angles.
Theor. 11.
As sin. of half sum of two sides	:
To sine of half their difference	::
So cotang. half the included angle	:
To tang. 1/2 diff. of the oppos. angles.

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Theor. 12.
As cosin. half sum of two angles	:
To cosine of half their difference	::
So tang. of half the included side	:
To tang. 1/2 sum of the opposite sides.
Theor. 13.
As sin. half sum of two angles	:
To sine of half their difference	::
So tang. half the included side	:
To tang. 1/2 diff. of the opposite sides.
Theor. 14. In a right-angled triangle,
As sin. sum of hypot and one side	:
To sin. of their difference	::
So radius squared	:
To square of tang. 1/2 contained angle.

Theor. 15. In any spherical triangle; The product of the sines of two sides and of the cosine of the included angle, added to the product of the cosines of those sides, is equal to the cosine of the third side; the radius being 1.

Theor. 16. In any spherical triangle; The product of the sines of two angles and of the cosine of the included side, minus the product of the cosines of those angles, is equal to the cosine of the third angle; the radius being 1.

By some or other of these theorems may all the cases of spherical triangles be resolved, both right angled and oblique: viz, the cases of right-angled triangles by the 1st and 2d theorems, and the oblique triangles by some of the other theorems.

In treatises on Trigonometry are to be found many other theorems, as well as synopses or tables of all the cases, with the theorem that is peculiar or proper to each. See the Introduction to my Mathematical Tables, p. 155 &c; or Robertson's Navigation, vol. 1, p. 162. See also Napier's Catholic or Universal Rule, in this Dictionary.

To the foregoing Theorems may be added the following synopsis of rules for resolving all the cases of plane and spherical triangles, under the title of

1. In a right-lined triangle, whose sides are A, B, C, and their opposite angles a, b, c; having given any three of these, of which one is a side; to find the rest.

Put s for the sine, s′ the cosine, t the tangent, and t′ the cotangent of an arch or angle, to the radius r; also L for a logarithm, and L′ its arithmetical complement. Then

. .

If A = B; we shall have	} C = ((s.(1/2)c)/r) X 2A.
a = b = 90° - (1/2)c, and

; from which equations any term wanted may be found.

When an angle, as a, is 90°, and A and C are given, then .

Note, When two sides A, B, and an angle a opposite to one of them, are given; if A be less than B, then b, c, C have each two values; otherwise, only one value.

II. In a spherical triangle, whose three sides are A, B, C, and their opposite angles a, b, c; any three of these six terms being given, to find the rest. |

.

And the same for the other angles.

And the same for the other sides.

Note. The sign > signifies greater than, and < less; also <01> the difference.

To find an angle a opposite the side A, .

Again let r : s′c :: t.A : t.M, like or unlike A as c is > or < 90°; and N = B <01> M.

Then s′M : s′N :: s′A, s′ C, like or unlike N as c is > or < 90°. Or, .

; . .

But if A be equal to B, or to its supplement, or between B and its supplement; then is b like to B: also c is = m ∓ n, and C = M ± N, as B is like or unlike a.

.

But if A be equal to B, or to its supplement, or between B and its supplement; then B is unlike b, and only the less values of N, n, are possible.

Note, When two sides A, B, and their opposite angles a, b, are known; the third side C, and its opposite angle c, are readily found thus: .

III. In a right-angled spheric triangle, where H is the hypotenuse, or side opposite the right angle, B, P the other two sides, and b, p their opposite angles; any two of these five terms being given, to find the rest; the cases, with their solutions, are as in the following Table. |

The same Table will also serve for the quadrantal triangle, or that which has one side = 90°, H being the angle opposite that side, B, P the other two angles, and b, p their opposite sides: observing, instead of H to take its supplement: and mutually change the terms like and unlike for each other where H is concerned.

Case	Given	Reqd	SOLUTIONS.
	H	b	s. H. : r :: s B : sb,	and is like B
1		p	r : t′H :: t. B : s′p	}, > or < 90° as H is like or unlike B
	B	P	s′B : r :: s′H : s′P
	H	B	r : s′H :: s.b : s.B,	like b
2		P	r : s′b :: t.H : t.P	}, > or < 90° as H is like or unlike b
	b	p	r : s′H :: t.b : t′p
	B	H	s.b : r :: s.B : s.H	}, each > or < 90°; both values true
3		P	r : t.B :: t′b : s.P
	b	p	s′B : r :: s′b : s.p
	B	H	r : t′B :: s′p : t′H,	> or < 90 as B is like or unlike p
4		b	r : s′B :: s.p : s′b,	like B
	p	p	r : s.B :: t.p : t.P,	like p
	B	H	r : s′B :: s′P : s′H,	> or < 90° as B is like or unlike P
5		b	r : s.P :: t′B : t′b,	like B
	P	p	r : s.B :: t′P : t′p,	like P
	b	H	r : t′b :: t′p : s′H,	> or < 90° as b is like or unlike p
6		B	s.p : r :: s′b : s′B	like b
	p	P	s.b : r :: s′p : s′P	like p

The following Propositions and Remarks, concerning Spherical Triangles, (selected and communicated by the reverend Nevil Maskelyne, D. D. Astronomer Royal, F. R. S.) will also render the calculation of them perspicuous, and free from ambiguity.

“1. A spherical triangle is equilateral, isoscelar, or scalene, according as it has its three angles all equal, or two of them equal, or all three unequal; and vice versa.

2. The greatest side is always opposite the greatest angle, and the smallest side opposite the smallest angle.

3. Any two sides taken together, are greater than the third.

4. If the three angles are all acute, or all right, or all obtuse; the three sides will be, accordingly, all less than 90°, or equal to 90°, or greater than 90°; and vice versa.

5. If from the three angles A, B, C, of a tri- angle ABC, as poles, there be described, upon the surface of the sphere, three arches of a great circle DE, DF, FE, forming by their intersections a new spherical triangle DEF; each side of the new triangle will be the supplement of the angle at its pole; and each angle of the same triangle, will be the supplement of the side opposite to it in the triangle ABC.

6. In any triangle ABC, or AbC, right angled in A, 1st, The angles at the hypotenuse are always of the same kind as their opposite sides; 2dly, The hypotenuse is less or greater than a quadrant, according as the sides including the right angle are of the same or different kinds; that is to say, according as these same sides are either both acute or both obtuse, or as one is acute and the other obtuse. And, vice versa, 1st, The sides including the right angle, are always of the same kind as their opposite angles: 2dly, The sides including the right angle will be of the same or different kinds, according as the hypotenuse is less or more than 90°; but one at least of them will be of 90°, if the hypotenuse is so.”