TRIGONOMETRY.

1. TRIGONOMETRY is a branch of the mathematics which treats of the relations of the sides and angles of triangles, these being expressed by numbers. It is therefore a theory formed by the union of arithmetic and geometry, and it treats of a particular class of geometrical figures, namely, triangles.

There are two kinds of triangles considered in trigonometry; plane triangles, and spherical triangles. The former are figures on a plane, and the latter on the surface of a sphere; hence we have plane trigonometry and spherical trigonometry, the principles of which will be explained in succession. In a triangle there are six things which may be considered, viz. the three sides, and the three angles. The main object of the theory is to give rules by which, when some of these are given, the others may be found.

2. The application of numbers to express the sides and angles of triangles has given rise to several terms which require to be defined before we proceed further.

DEFINITIONS.

I. Since angles at the centres of equal circles have to each other the same ratio as the arcs on which they stand, the arcs may be taken as representatives or measures of the angles to which they are proportional. Thus, let ACB be any angle contained by the lines AC, BC: if about C as a centre, with any distance or radius AC, a circle ABE be described, the arc AB, intercepted between the lines CA, CB, is the measure of the angle ACB.

II. The circumference of every circle is supposed to be divided into 360 equal parts called degrees, and again each degree is subdivided into 60 equal parts called minutes, and each minute into 60 equal parts called seconds; and an angle is said to be of as many degrees, minutes, seconds, &c. as there are degrees, and minutes, and seconds in its measure. The degrees in any arc or angle are indicated by the character o placed over their number, the minutes by an accent ', and the seconds by two accents ". For example, an angle of 25 degrees 12 minutes and 14 seconds is expressed thus, 25° 12' 14''—A right angle is measured by one fourth part of the circumference, or a quadrant, and it contains 90°.

III. The complement of an arc or angle is the difference between it and a quadrant, or 90°. Thus, supposing AD to be a quadrant, the arc BD is the complement of the arc AB, and the angle BCD is the complement of the angle ACB.

IV. The supplement of an arc or angle is its difference from half the circumference, or 180°. Let ADE be a semicircle; the arc BDE is the supplement of the arc AB; and the angle BCE is the supplement of the angle ACB.

V. The sine of an arc or angle is the perpendicular drawn from one extremity of the arc, on the diameter, that passes through the other extremity. The line BF is the sine of the arc AB, or of the angle ACB.

VI. The versed sine is the segment of the diameter between its sine and the arc which measures the angle. The line AF is the versed sine of the arc AB, or the angle ACB, and EF the versed sine of the angle BCE.

VII. The tangent is a straight line that touches the arc at one end, and is limited at the other end by a straight line drawn from the centre of the circle through the other extremity of the arc. The line AH is the tangent of the arc AB, or the angle ACB.

VIII. The secant is the straight line drawn from the centre to the extremity of the tangent. The line CH is the secant of the arc AB, or angle ACB.

IX. The sine, tangent, and secant, of the complement of any arc or angle, are called by abbreviation its cosine , cotangent, and cosecant. Thus, BG = CF, the sine of the arc BD, is the cosine of the arc AB; and DK, the tangent of the arc BD, is the cotangent of AB; and CK, the secant of BD, is the cosecant of AB.

Trigonometrical Notation.

The sine of an angle whose measure is A, is expressed thus, sin. A ; its cosine by cos. A ; its tangent by tan. A; its cotangent by cot. A; its secant by sec. A; its cosecant by cosec. A; its versed sine by ver. siri. A; and its coversed sine by co. ver. sin. A.

The radius of a circle, when it is supposed the unit of numbers, requires no symbol: sometimes it is represented by the abbreviation rad. and also by the letter R.

3. There are two ways in which the theories of trigonometry have been established. One is by employing a process of reasoning purely geometrical, such as has been followed in our article Geometry. In this way the early writers on the subject constructed their treatises, and it served very well for plane triangles, but was less convenient for spherical triangles, by reason of the intricate diagrams required to represent circles lying in different planes; and also by the complicated relations of the things to be discussed, which could not easily be expressed by words. The second, which is the more convenient way, is by the symbolic reasoning of algebra. The section in Algebra on the Arithmetic of Sines is a specimen of this way of treating the subject; and it is worthy of remark, that the great extension that has been given to the subject is in no small degree due to the simple and convenient way of representing the sines and tangents, &c. of angles, by the abbreviations, sin. A, tan. A, &c. instead of referring to the geometrical lines themselves. As a consequence of this, geometrical constructions are avoided, and even diagrams are hardly necessary.

4. In applying the notation and doctrines of modern analysis to trigonometry, we regard the measures of angles, and their sines, tangents, &c. as endowed with the properties of variable quantity; in particular, we ascribe to them the quality of being positive or negative. The consequences of this hypothesis have been fully detailed in the Arithmetic of Sines (Algebra), to which we refer the reader. By this the properties of triangles are expressed by general formulae, which comprehend all particular cases. This way of considering the numeral measures of trigonometrical lines has greatly simplified spherical trigonometry, and all applications of both trigonometries to astronomy and geodetical inquiries.

The Trigonometrical Table.

5. The basis of all trigonometrical calculations is a table which exhibits in parts of the radius the numeral values of the sine, tangent, and secant, of each arc in a series pro [7:21:371]ceeding from 0 to a quadrant, the common difference of the terms being an angle of one minute, or some smaller angle.

The radius being considered as an unit, the sines are expressed as decimal fractions of that unit. The quadrant is generally divided into degrees and minutes, this being sufficient for most purposes; but tables have been calculated which give the sines, and tangents, and secants, to every tenth second.

6. The Elements of Euclid show how to inscribe regular polygons of three, four, five, six, and fifteen sides in a circle. The sides of these polygons may be expressed in numbers by calculations derived from the geometrical constructions. Now, the side of an equilateral triangle in a circle, is the chord of 120°, which is twice the sine of 60°; and the side of the inscribed square is the chord of 90°, which is twice the sine of 45°; and the side of the equilateral pentagon is the chord of 72°, which is twice the sine of 36°; and the side of the hexagon is the chord of 60°, which is twice the sine of 30°; and the side of the quindecagon is the chord of 24°, which is twice the sine of 12°. These sines may therefore be all deduced from the properties of polygons inscribed in a circle, as delivered in the Elements. And since the sum of the squares of the chord of an arc and the chord of its supplement is equal to the square of the diameter (Geom. 17 of 2 and 13 of 4), the chords of the supplements, and thence the sines of their halves, may be found. It was in this way the trigonometrical table was first constructed by the early mathematicians, Purbach, Miiller, Copernicus, G. Joachimus Rhaeticus, and others. The theory of inscribed polygons was extended by Vieta into that of angular sections, chiefly with a view to the construction of the table. This has again been wrought up into the Arithmetic or calculus of Sines, the formulae of which afford great facility in the construction of the table. As an example, we shall resolve this problem.

7. Problem. To find the sines of one fifth, two fifths, three fifths, and four fifths, of a quadrant, of which the sine is equal to the radius. Putting a to denote an arc of 18°, one fifth of the quadrant, we are to find the sines of a , 2 a, 3 a , 4 a . Now, from the nature of the problem, we have these simple relations:

*******Cos. a = sin. 4 a , cos. 2α = sin. 3«,

Cos. 3« = sin. 2a, cos. 4a = sin. a.

And by formula 3 of (E) art. 240, Algebra,

Sin. 2α = 2 sin. a cos. α,

Sin. 3a — 2 sin. 2a cos. a — sin. a,

Sin. 4a = 2 sin. 3a cos. a — sin. 2σ,

Sin. 5a = 2 sin. 4a cos. a — sin. 3«.

Therefore Sin. 2σ = 2 sin. a sin. 4a,

Sin. 3a = 2 sin. 2a sin. 4a — sin. a,

Sin. 4a = 2 sin. 3a sin. 4a — sin. 2zz,

Sin. 5⅛ = 1 = 2 sin. 4a sin. 4a — sin. 3a.

To abridge, put sin. a = ιι, sin. 2a = x, sin. 3a — y, sin. 4a = r, and we have the following equations:

2«z = X A, 2xz = y + u B, 2gz = z + X C,

2zz = 1 + y................ D.

These express symmetrical relations, which connect the four sines w, x, y, z.

By subtracting A from C, and dividing the remainder by 2z, we have y —« = J (i).

And from B, A, D; y + u = 2xz = 4ι<a* = 2a + 2ιry. Therefore 2uy = y — « = ⅜; 5 and hence, (y + w) 2 = (y— u) 2 + 4uy =

and y + u = yt (2).

We have now, by subtraction and addition from (1) and (2),

V5 — 1. v ’ 5-4-l sin. α — w — — , sin. 3a = y — ÷ j —.

+ ʹ r

Now' the arcs a and 4o being the complements of each other, also 3a and 2a, and the sum of the squares of the sine and cosine being equal to the square of the radius, we have ein. 2« = « = √(1 — y ’) - √(10-∙g√¾), ⅛,⅛=,⅛√(l-⅛) = << ll, + i ^>

Our problem is now resolved.

8. We have found the sines of a series of arcs from 0 to 90°, viz. 18°, 36°, 54°, 72°, whichdiffer by 18°. From these, the sines of a series whose terms differ by 9°, may be found by formulae, which are modifications of formulae (1), (2) of (P), article 249, Algebra, viz.

sin. I (90° — a) = t f- ~∙^ ln ' ", sin. ⅜ (90° + α) =

The terms intermediate to those known are,

. no /1 —sin.72° . /1 +sin. 36°

sm. 9° = J g > sln ∙ 63 = √ ÿ , o*o— /’ — sin. 36° . ∕l÷βin. 72“

sin. 2.° = J , sin. 81° = J , sin. 45° = yi.

By a second interpolation the number of terms in the series would be doubled, and by a third, the sines of a series of arcs which are multiples of 2° 15' would be found. In the same way, from sin. 30° = ½, sin. 60° = — , and sin. 90° = I, we may deduce the sines of 15°, 45°, 75°; and at length the sines of a series of arcs, which are multiples of 3° 45'; then, from the two series, another, the arcs of which would differ by (3° 45') — (2° 15') = 1° 30' = 90'; and again the sines of a series of arcs differing by 45'. Indeed it was in this way the table was first formed.

9. When the sines of two thirds of the quadrant, that is, from 0 to 60°, are known, the sines of the remaining third can be readily found; for since sin. (a + b ) = sin. (a — b ) + 2 cos. a sin. b (Algebra, art. 240), if we make a = 60°, then cos. a = ½, and 2 cos. a = 1, and we have sin. (60° + b) — sin. (60° — b) + sin. b. By this formula the numbers in the table may be verified.

The radius being an unit, we have found that the semi-circumference or arc of 180° = 3·1415926535 (Algebra, article 272). This number divided by 180 × 60 = 1080, gives the arc which measures an angle of one minute = ·0002908882; and since a small arc is nearly equal to its sine, we have sin. 1'=∙0002908882 nearly. The accuracy of this may be verified by the formula ******Qp cP sin. a = a — — + —- —, &c. (Algebra, art. 269).

The cosine of one minute is found from the sine by the formula cos. α = √(1 — sin. 2 a), or otherwise, by the series cos. a — 1 — —, &c. ≡ ∙9999999577. Knowing now the sine and cosine of one minute, to find the sines of 2', 3', 4’, &c. we may proceed thus,—

Sin. 2' — 2 cos. 1' sin. Γ,

Sin. 3' = 2 cos. 1' sin. 2' — sin. P, Sin. 4' = 2 cos. 1' sin. 3'—sin. 2 ’, &c.

10. Since the results obtained by the successive operations[7:21:372] are only approximations, they will diminish in accuracy as we advance. We may however proceed otherwise thus. Find, as has been explained, a series of sines of arcs differing by 45' from 0° to 60°; then fill up, by interpolation, the intervals thus: We have found (Algebra, art. 239) *********sin. (a + l>) + sin. (α — ⅛) = 2 .cos. b sin. a: instead of a write x + h, and instead of 6 write h, and we have sin. lx + 2h) + sin. x = 2 cos. h sin. (* + A).

, , A s , P h β

Now cos. Ä = 1 — τ - 5 + — — -.- j-.∙. γ - g» &c ·

Γ2 l∙2∙3∙4 l∙2∙o∙4∙5∙o h* h 6

Putn to denote the series h- —∙ - + - , —-, — &c. and U*T m‘4'0’0 we have sin. (.1· + 2A) + sin. x = 2 sin. (.r + h) — n sin. (x + A); and sin. lx + 2Λ) = sin. (r + A) + (sin. (r + A) —sin.a:) — n sin. lx ⅛ A). If A be an arc of one minute, then A = ∙0002908882, and n — ∙t)000000846. Let x denote any angle, and suppose A to be one minute; the formula gives sin. (a: + l') = sin. x + (sin. x — sin. lx — Γ))

— n sin. X, sin. lx + 2') = sin. lx + Γ) + (sin. lx + l') — sin.

— n sin. (a· + l').

sin. lx + 3') =sin. (® + 2') + (sin. lx+ i 2')— sin.(≈+ Γ))

— n sin. lx + 2').

Thus, from the sines of two arcs x— 1' and x , which differ by 1', and the constant number n = 2 ( 1 —cos. l') = (2 sin. 30'') 2 , the sines of all angles or arcs exceeding x, and differing by l', throughout the table, may be found. The most extensive table of sines in existence (which is yet in manuscript), was computed under the direction of the late Μ. Prony in this way. It gives the sines of all arcs in a series of which the common difference is one ten-thousandth part of a quadrant with twenty-five decimals.

11. The sines being found, the tangents and secants may be obtained, each by a single division from the formulae tan. a = sin. a / cos. a , sec. a = 1 / cos. a .

The tangents of the latter half of the quadrant may however be found by addition only from the formula tan. (45° + a) = tan. (45° — a ) + 2 tan. 2rr, which is a deduction from (4) of (O) Algebra, art. 248; and the secants from this formula, sec. a = tan. a 4- tan. (45° — ½ a ), Algebra, 249.

12. With a table of sines, and tangents, and secants, the ordinary questions relating to triangles may be resolved by multiplication and division, but more expeditiously by logarithmic calculation; and therefore, in general, mathematical tables give only the logarithmic sines, tangents, and sometimes secants, which, however, may be readily found from the log. cosines.

Part I.— Plane Trigonometry.

13. In contemplating a plane triangle as a figure formed by three straight lines, we distinguish the sides of the triangle, the angles at their intersections, and the space or area which they comprehend. In trigonometry we abstract from the area, and consider only the sides and angles. There are in all six angles, viz. the three interior and their adjacent exterior angles. Each of the exterior is the sum of two interior angles, and of these any two determine the third; so that in fact the independent elements of a triangle are five, viz. its three sides, and any two of its three interior angles; and of these any three being given, the remaining two may be found.

14. The sides and angles are dissimilar in their nature, and therefore do not admit of being directly compared: the measures of angles, being arcs of a circle described with a given radius, are lines, and therefore expressible by the sides. It is however more convenient and easy to use the sines and tangents, &c. of angles than the angles themselves, in trigonometrical calculations, the rules of which are to be deduced from the following theorems.

***********Τηεοrεμ I. In any right-

angled triangle, the radius is to the sine of either of the acute angles as the hypotenuse is to the side opposite that angle. (Fig· 2.)

Let ABC be a right-angled triangle, of which B is the right angle; about C, one of the acute angles, as a centre, with a radius equal to the radius in the tables, describe an arc DE as the measure of the angle C (Der. 1). Draw DP, the sine of the arc DE (Der. 5). The triangles CDF, CAB are manifestly equiangular, therefore their sides are proportionals; so that CD: DF ≡ CA: AB, or (putting R to denote radius in the tables) R: sin. C = CA: AB.

Corollary 1. The radius is to the cosine of either of the oblique angles as the hypotenuse is to the side adjacent to that angle, l∙ , or. to the radius CD, CP is the cosine of C, and CD: CF = CA: CB.

CorMiu∕ 2. Assuming that CD = R = 1, we have

1: sin.' C = CA: A B: and 1: cos. C = CA: CB. Therefore, suj>posing the sides of the triangle expressed by numbers,

. r , AB CB

sui. C = ,-, and cos. C = _ ..

CA CA

***********THEOREM II. In any right-angled triangle, the radius is to the tangent of either of the oblique angles as the adjacent side to the opposite side. (Fig. 2.)

At the point E, one extremity of the arc DE which measures the angle C, draw EG perpendicular to CB; then to the radius CE, EG is the tangent of the angle C (Der. 7). Now the triaiig!es CEG, ABC being similar, CE: EG = CB: BA, or R: tan. C = CB: BA.

corollary 1. The radius is to the secant of either of the oblique angles as the adjacent side to the hypotenuse; for CG is the secant of the angle C (Der. 8); and CE: CG = CB: CΛ, or R: sec. C = CB: CA.

corollary 2. Assuming the radius = 1, then

. n V BA a r> CA tan. A = —, and sec. C = —.

tv 15 L 1>

**********f οrollary 3. In any triangle, if a perpendicular be drawn from any one of the angles to t)>e opposite side, the segments of that side are to one another as the tangents of the parts into which the opposite angle is divided by the perpendicular. r

For in the triangles ADB, ADC,

BD: DA = tan. BAD: R; and DA: DC = R: tan. CAD. Therefore, ex aquali, BD: DC — tan. BAD; tan. CAD.
[7:21:373]

Theorem III. In any right-angled triangle, the sum of the hypotenuse and one side is to their difference, as thç square of the radius to ψe s<piare of the tangent of half the contained angle.

Let ABC be any right-angled triangle; draw CD bisecting the angle C. By Geom. 19, 4,

CAtCB = AD :BD = AB —BD: BD.

Therefore, by composition, and Cor. 3 of Theor. II. of Tr!g.

C A + CB: CB ≡ A B: BD = tan. C: tan. ’ C. Now CB: BA = R: tan. C, therefore CA 4- CB: BA = R: tan. ⅞C, and (CA -fCB) ’: BA* = R-: tan. f AC·

Now BA s - CA∙ i - CB e = (CA 4- CB) (CA — CB) (Gκr>M. 9 and 18 of 4), therefore

(CA + CB)’: (CA CB) (CA — CB) = R-: tanA∣C, and C A + CB: C A — CB = R»: tan.« ⅜(!.

The first two of these pronositions are alone sufficient for the resolution of all the cases of right-angled triangles; the third, which is not commonly given in our books on trigonometry, is however im∣>ortant, because when the lrypo-tennse and a side are given, by its application the same logarithms serve for finding the remaining side and the tangent of half the opposite angle.

15. It has been found conducive to brevity and perspicuity, to represent the parts of a triangle, each by a single letter: therefore, in any triangle ABC, we shall, in what follow«, employ that notation to denote the sides and angles as follows. (See figures 2, 3, 4.)

The sides, BC =α, AC = b, AB = c; the angles, BAC = A, ABC = B, ACB = C.

It will also be sometimes convenient to express ratios by fractions, the numerators of which are the antecedents, and the denominators the consequents, so that the expressions A C

A: B = C: D, and ⅛r = w∙, arc to be regarded as identical. B 1J

***THEOREM IV. The sides of any plane triangle are to one another as the sines of the opposite angles. ( Figs. 3 and 4.) In the triangle ABC, draw AD perpendicular to BC; then, employing the premised notation, that is, denoting the sides by ♯, b, c, and opposite angles by A, B, C, b: AD = R: sin. C; and AD: e = sin. B: R; therefore, ex aquali, inversely, b: c = sin. B: sin. C.

***THEOREM V. The sum of the sides of any plane triangle is to the difference, as the tangent of half the sum of the opposite angles is to the tangent of half their difference.

Continuing the same notation, we have found that b: c = sin. B: sin. C; therefore b 4- c: b — o = sin. B -fsin. C: sin. B — sin. C. Now (Algebka, 247, M), sin. B -∣- sin.C: sin.B—sin.C=tan. A (B 4- C): tan. ∣ (B—C); therefore, b 4- c: b — c=, tan. ⅛ (13 4- C): tan. -∣ (B — C). corollary. Since B 4- C = 180°— A, and therefore ⅛(B + C) = 90°-⅜A, therefore b -f- c: b — c= cot. ∣ A: tan. J (B — C).

***THEOREM VI. In any triangle, the cosine of half the difference of any two of its angles is to the cosine of half their sum, as the sum of the sides opposite these angles is to the remaining side. And the sine of half the difference of the angles is to the sine of half their sum, as the difference of the opposite sides to the remaining side.

Again, denoting the sides by a, b, c, and opposite angles by A, B, c,

Because sin. A: sin. C = α: c, and sin. B: sin. C == b: c; therefore sin. A + sin. B: sin. C '= a -f- b : c, also sin. A — sin. B: sin. C = α — b i c.

Now, sin. A + sin. B = 2 sin. 4 (A + B) cos. A (A — B) (Algebra, 240), and sin. A — sin. B = 2 cos. ⅜ (A + B) sin. ∣ (A — B), and sin.C= sin.(A + B) = 2sin. * (A-f-B)cos.∣(A-t-B) (Algebra, 245); therefore sin. A + sin. B; sin. C = cos. ⅛ (A — B): cos. ⅜(A + B), and sin. A — sin. B: sin. C ~ sin. ⅜ (A — B): sin. i (A + B); hence cos. A (A — B): c∙os.^⅜ (A 4- B) = a *+ b: c, and sin. Î (A — R): sin. ⅜ (A -jB) = α — b: c; ***Theorem VII. If a perpendicular be drawn from an angle of a triangle on the opposite side, dividing it into two segments, the sum of these segments is to the sum of the sides as the difference of the sides tq the difference of the segments.

In the triangle ACB, AC* = AD 2 4- CD 5 , and AB 2 = AD- i + DB 2 , therefore AC’ — AB= = CD i — BD-. Now AC 2 — AB 2 = (AC + AB)(AC-rAB), and CD'— BD 2 = (CD + BD) (CD — BD) (Geometry, 12, 4); therefore (AC+AB) (AC— AB)-(CD÷BD)(CD-BD), and CD + BD: AC + AB = AC — AB: CD — BD.

***TheoreM VIII. In any plane triangle, the excess of the sum of the squares of two sides which contain an angle above the square of the third side, is to twice the rectangle contained by the sides about the angle, as the cosine of the angle is to radius. (Figs. 3 and 4.)

In the triangle ABC, draw AD perpendicular to BC, we have, by Geometry, 14, 4, and Cor. 2, Theorem I., of this, c- = a* + lf— 2α X CD, and CD - .

therefore c t — a" + b i — 2ab —, 1 α s + b° — e ’ cos, C and ~⅞⅛ = ~R~ ι therefore a° + br — c°: 2ab — cos. C: R.

We have, in the diagram, made the angle C acute, but the proposition is equally true when C is an obtuse angle; because, when C changes from acute to obtuse, the sign of the cosine of the angle changes from -fto —.

***Theorem IX. In any triangle, the rectangle contained by two sides is to the rectangle contained by half the perimeter and the excess of half the perimeter above the base, as the square of the radius to the square of the cosine of half the angle contained by the sides.

It has been found that a, b, c, being the sides of a triangle, and A the angle contained by b and c, then, Theorem VI IL, cos. A _ b i 4- e 2 — α'

- R^" - 2⅛c '

To the first of these equals add ⅛ 5 = 1, and to the second, κ

20o

‰= 15 then

R 4- cos. A _ δ 2 4- 20c-∣-c 2 — α 3 _ (ά + o) 2 — α 2

IÎ - 26c - 2⅛ [7:21:374] ***.. R 4-cos. A 2 cos.* ⅜ A z √>< o D\ i

Now —ö = „ a " (Algebra, 249, P), and

(6 + c) t — a" ≡ (α + b + c) (b + c — a).

Put s = ⅛(a + b + c), then b + c — a = a + b -fc—2a = 2(s — a) ; we have now (b + c) 2 — a 2 = 4s(s — a), and, i . cos.' A A s(s — a) . by substituting, —⅜⅝ = and be ∙. s(s — α) = R': cos.’ ∣A.

***Theorem X. In any triangle, the rectangle contained by two sides is to the rectangle contained by the excesses of half its perimeter above these sides, as the square of the radius to the square of the sine of half the angle contained by the sides.

Supposing, as in last proposition, that b and c are the sides about an angle A, and a the side opposite to a, we have, by Theorem IX.

cos. A _ b° + c* — α 2

R 2bc

Subtract the first of these equals from — 1, and the se-

26c cond from -- τ - ~ 1, and we have

26c

R — cos. A _ 2bc — 7r 2 — c' + α ’ a- — (b — c) 2

R 26c 20c

.. —cos.A 2sin. s ⅜A..

Now g ≡ — r 8 8 (Algebra, 249, P), and a^ — (b — c) 2 = (α + b — c) (a — b + c).

Put s = ∣(a + b + c), then a + b — c = a + b + c — 2c = 2s — 2c, and a — 0-f-c=a4-Z> + c — 2b = 2s — 2b; we have now α t — (b — c) i = 4(i — b) (s — c), and gin ∙* - <* — 6 ) (* — c >

R 2 - be ’ and be : (s — b) (s — c) = R ’: sin.* ∣ A.

***THEOREM XI. In any triangle, the rectangle contained by half the perimeter, and its excess above the base, is to the rectangle contained by the excesses of half the perimeter above the two sides, as the square of the radius to the square of the tangent of half the angle contained by the sides.

As in the two preceding theorems, putting b and c for the sides about the angle A, and a for the opposite side, and s =∣ (α -∣- b -∣- c), we have found that s (s — a) ∙. be = cos. 2 A: R", and be: (s — b) (s — c) = R': sin.'∣A; therefore, ex tεquali, s(st-a): (s — b)(s — c) = cos.' ⅛A: sin.’ ∣A.

But cos.’ ∣A: sin.’ LA = R'; tan.' $ A; therefore s(s — α): (s — b) (s — c') = R 2 : tan.’ ∣A. Scholium. Since we have tan.'|A_(s— b)(s —c)_ 1 (s— α)(s — b)(s—c)

R' “ »(* —α) - (4 - θ )' - s ’ let us put M to denote the expression — π ∖(* ¾*∑τ5),

X s and we have — = ∙ have like expressions for the angles B, C, and, on the whole, tan.∣A = — R, tan.⅜B = R, tan JC = — R.

s — a ∙ i s b * s c

This seems to be the most convenient formula for finding the angles of a triangle when the sides are given.

16. Having laid down principles, we are next to apply them to trigonometrical calculation. The foundation, as has been already observed, is a table which shows, in parts of the radius, the lengths of the sine, and tangent, and secant, to every minute or smaller division of the quadrant. The most extensive table of this kind is that of G. Joachimus Rhaeticus, published in 1596. It gives the sines, &c. to every ten seconds. The invention of logarithms has in a great measure superseded the use of the natural sines, as they are called, and, except for particular purposes, the logarithms of the natural sines, called logarithmic sines, only are used. Of tables containing these, there are innumerable varieties, according as they are of greater or less extent. The Arithmetica Logarithmica of Briggs, published in 1628, by Adrian Viacq, in two folio volumes, is the most extensive. His tables give the logarithms of 100,000 numbers, and logarithmic sines and tangents to every ten seconds and to ten places of decimals. Hutton’s Tables give the natural sines, and tangents, and secants; also their logarithms to every minute and to seven figures. Callet’s Tables Portatives give the logarithmic sines and tangents to every ten seconds; and Taylor’s Tables, in a large quarto, give the same to every second. For general use, Callet’s are the most convenient. Tables which give the logarithms of numbers and sines, &c. to six, and even to five figures, are also useful. Of the former, we recommend Tables of six-figure Logarithms, superintended by Richard Farley of the Nautical Almanac Office (printed for Longman and Co. 1840); and of the latter, Tables of Logarithms, under the superintendence of the Society for the Diffusion of Useful Knowledge (Taylor and Walton, 1839). Those who perform many calculations find it convenient to have tables of different extent. The size of our work, and the form of its pages, make it quite unsuitable for a useful logarithmic table, which should be as portable as it can be made. Accordingly we have given none.

17. It is customary to treat first of right-angled triangles, and next of such as have all their angles oblique.

RIGHT-ANGLED PLANE TRIANGLES.

These form four distinct cases. Adhering to the notation premised in article 15, we shall denote the hypotenuse by the letter b, the sides about the right angle by a and c, and the opposite angles by A and C. (See fig. 2.)

***Case I. Given the side a, and either of the acute angles A, C, and consequently (Geometry, 24, 1) both; to find the hypotenuse b and remaining side c. Solution. R: tan. C = side a : side c;

R: sec. C = side a : hyp. b : or cos. C: R = side a : hyp. b. Case II. Given the hypotenuse b, and either of the acute angles A, C; to find the sides a, c about the right angle B.

Solution. R: cos. C = hyp. b : side a,

R: sin. C = hyp. b : side c.

Case III. Given the sides a , c about the right angle B; to find the angles A, C, and hypotenuse b. Solution. Side a : side c = R: tan. C, cos. C: R = side a : hyp. b.

The hypotenuse C may be found independently of the angles by the formula b = √(n' -fc') (Geometry, 13, 4). The angle A is found from C, as above.

Case IV. Given the hypotenuse b, and a side u; to find the angles A, C, and side c.

Solution. Hyp. b : side a = R: cos. C; or thus, b 4- a: b — a = R': tan.’ ∣C; then side c = √ ((δ -fα) (b— β)].

The advantage of this solution is, that the same logarithms, viz. of b -f- α and b — a, serve to find the side c and angle C.
[7:21:375]

18. Examples of the solution of right-angled triangles.

Case I. Given-f on ⅛θp - r√ 1 V44∕ Required side c

1 angle Ç = 55 44, , and 1 1 t . 6 .

and therefore angle A = 34 16. J jt calculation by Logarithms-

R 10-00000 Cos. C 9∙75054

Tan C lth∙16666 R 10∙006(M)

Side a 2∙7lO96 Sidea 2∙7l096

12∙87762 12∙71096

Sidec = 754∙4 2∙87762 Hyp fc = 912∙9 2∙96042

Here we add together the logarithms of the second and third terms of each proportion, and subtract the logarithm of the first from their sum. The remainder is the logarithm of the fourth term, which is the answer.

Case II. Given ( ⅛ * = ⅛~ ’ 12 ,, I Required sides and therefore angle A = 47 48. ) α and t '

R 10-00000 R 10-00000

Sin. A 9∙86970 Sin. C 9∙82719

Hyp. b 2∙655l4 Hyp. b 2∙65514

Side « = 334-9 2∙52484 Sidec= 303∙6 2∙48233

To save figures, the logarithm of radius may be subtracted or added mentally.

r , τrτ r~∙∙ f side a = 4681 Required angles A

***Case III. G>ven; gin c = 305 ∣ and ∣ 1 c , and hyp. b.

Side α 2∙67025 Cos. C 9∙89683

Side c 2∙56229 R 10 00000

R 10-00000 Side« 2∙67025

Tan. C = 37° 57' 9∙89204 Hyp. b = 593∙5 2∙77342

The hypotenuse b may also be found, by extraction of the square root, thus:

5 = √(α s +c a ) = √(468*+365 2 )= √(3522409) =593∙5.

***Case IV. Given ( ⅛P∙ * = 664Λ r

1 ., r I b + a = 978? A, C, and side c.

and therefore ∣ 0 22 fl = 350 J

Hyp. b 2∙822l7 R 10-00000

Side a 2∙49693 Sin. C 9∙94506

R 10-00000 Hyp. b 2∙82217

Sin. A = 28° 13∣' 9∙67476 Side c 585∙1 2∙76723

This case may be resolved also thus, by Theor. III.

b 4. a...................... 2∙99034 b 4- a 2∙99034

b — a 2∙54407 b-α 2∙54407

R 2 20-00000

2)5∙53441

Tan.’ iC 2)19∙55373

c= 585∙1 2∙76720

Tan. I C = 30° 53⅜' 9∙77686

C = 61 46∣

This second solution, although not given in our books of trigonometry, is better than the first, because the two things sought are obtained by the same logarithms, and independently of each other.

No notice is here taken of geometrical constructions of the cases by scales. We however recommend that the student construct all the cases geometrically, and refer him to the article Navigation for examples and practical directions.

OBLIQUE-ANGLED TRIANGLES.

19. The three angles of any plane triangle being equal to two right angles (Geometry, 24, 1), when two of them are given, the third is also given.

There are four cases of oblique-angled triangles.

Case I. When a side a, and two angles B, C, and consequently the third angle A, which is the supplement of B + C, are given; to find the sides b, c. ***SolutION. Sin. A: sin. B =α: bl , π ...

Sin. A: sin. C=a: c∫ 11, ∙ * ' ∙

Case II. Two sides b, c, and an angle C opposite to one of them, are given; to find the other two angles A, B, and the remaining side a. (Fig. 7.)

SolutION. Side c: side b — sin. C: sin. B (Th. IV.) When B is found, A = 180° — (B 4- C) is known; and sin. C: sin. A ≡ side c: side a.

Since any two angles, which make together two right angles, have the same sine, and the angle B in this case is to be found by its sine only being known; it will have two distinct values, viz. B, and B'= 180°—B. The angle A may also have two values, A = 180°—(B + C), and A'= 180° — (B' + C). Thus there will be two triangles ABC, AB'C, which alike satisfy the conditions of the data. It must however be considered that the three angles of a triangle cannot exceed two right angles, therefore if C be an obtuse angle, B can only be an acute angle, because a triangle cannot have two obtuse angles. Farther, if the side c opposite to the given angle C be greater than b, in this case B can have but one value; for when c is greater than b, then C is greater than either of the values of B, one of which is 180 — B (Geometry, 13, 1), and C + B will be greater than 180°, which is impossible (24, 1).

***Because 7- = -7—⅛, and sin. B cannot exceed the radius, b sin. B

therefore c τ cannot be less than -∙⅛∙ -; and unless the data b . it satisfy this condition, the solution will be impossible.

Case II. Two sides b, c, and the angle A between them, are given; to find the angles B, C, and the remaining side a. V

***SolutioN. When A is given, B + C its supplement is given; then b 4- c ∙.b—c — tan. ⅜(B + C): tan. <j(B— C). (Theorem V.)

Thus A(B — C) becomes known, and B = ∣(B -∣-C) 4- £(B — C), also C = ∣(B + C) — J(B — C), are known. All the angles of the triangle being known, and two of its sides; the remaining side may be found in two ways by Case I., or better by either of these proportions,

Cos. ⅛(B — C): cos. ⅛(B -∣- C) = b + c∙. a, l y∣ y; Sin.⅛(BC)rεin.4(B + C) =b — c-.a f lheor ∙ Vk If both be used, the results will verify each other. The side a may also be found by Theorem III.

Case IV. The three sides α, b, c are given; to find the three angles.

***SolutioN. Find s = ∣(α + b 4- c), andM = √(<-

M

Then, by Theorem XL, tan. ⅛A — — — ∙ R, tan. ⅛B

= ∙ R, tan. ⅜C = —— ∙ R.

9 — b * 9— c [7:21:376]By this rule, the values of the three, angles may be readily found, and also a verification of the calculation, because the sum of the halves of the angles must make ninety degrees.·

Other solutions may be had from Theorems VII. VIII. IX. X. That from Theorem VII. (which resolves the triangle into two right-angled triangles by a perpendicular on one of its sides from the opposite angle) is very generally used by practical men not much versed in theory; but the best solutions are those which follow from Theorems IX. X. XL These elegant formulae were first found about the year 1653, by William Purser of Dublin. The modification of Theorem XL, given here as a practical rule, is the best method we know for resolving this case.

***Examples of Oblique-angled Triangles. f side a = 575; ***Case I. Given angle B = 63° 48’ ( Required the

( angle C = 49 25 f sides Z>, c. and therefore angle A = 66 47 )

Sin. A 9∙9C>333 Sin. A 9∙9f>333

Sin. B 9∙95292 Sin. C 9∙8805l

Side a 2∙75967 Side a 2∙75967

12-71259 12-64018

Side b = 56l∙4 2∙74926 Side c 475∙2 2’67685

***Case II. Given P⅛ e 6 = ) Required an-

fFiB∙7l ι s,d '' c = 2i2 .∙ glcs A, B, and

< rι 8 ’ '*' ( angle C = 37° 20') side a.

Side c 2∙36549 From 180° 0'

Side b 2∙53782 subtract B -∣- C — 101 44

Sin. C 9∙78280

angle CAB = 78 16 12-32062

From 180 0

Sin.B= 64°∙2U I 9 . 9 - 51 subtract B’-f-C = 152 56 or sin. B'= 115 36 ∫ , angle CAB' = 27 4

In this example the angle A has two distinct values, from which we shall find two values of BC = a, and B'C = a'.

***To find BC. To find B'C.

Sin. C 9-78280 Sin. C 9∙78280

Sin. BAC 9∙99083 Sin. B’AC 9∙65S04

Sidec 2∙36549 Side c 2*36549

12-35632 12-02353

Side BC = o....2∙57352 Side B'C = β'....2∙24073

= 374∙6 ≡ 174∙ I

***Example 2. In a plane triangle, there are given two sides, 532 and 358, and the angle opposite to the former, 107° 40'; to find the remaining angles and side.

In this example, the things required have each only one value. The angles will be found 39° 53' and 32° 27', and the remaining side 299∙6.

( side b — 1230 ] Requiret! angles ***Case III. Given - sido c =x 879 ,-B, C, and side ( angle A ≡ 105°) a. b + c = 2100, b — c - 360, B -fC = 180° — A = 75°, J,(B 4- C) = 37° 30'

b + c 3∙32222 Cos. ( (B — C) 9∙99628

b — c 2∙5563O Cos. £ — (B + C) 9∙89947

Tan.j(B + C) = 37° 30' 9∙88498 b+e 8’32222

12-44128 13-22169

Tan.⅜(B-C)= 7 29⅛9∙ll906 «= 1680-4 8’22541

B = 44 59∣ C = 30 0∣

As a verification, a may be found from this other proportion:

Sin. ½ (B — C): sin. ½ (B + C) = b — c : a.

The side a may also be found by Case I. in two ways, because all the angles and two sides are known.

20. In all the preceding operations, the logarithm of the fourth term of each proportion has been found by subtracting the logarithm of the first term from the sum of the logarithms of the second and third. Now, in every case, when one number a is to be subtracted from another number b, we may obtain the result required by adding the difference between a , and n any number, and then subtracting the number n from the result. Thus, instead of subtracting 7 from 12, we may add 3, the difference between 7 and 10, to 12, and then rejecting 10 from 15, the sum, obtain 5 for the difference between 12 and 7. The reason of this may be easily understood, and it is evident from algebraic subtraction; for A — a — b + (n — a ) — n. When a , the number to be subtracted, is less than 10, then n is conveniently assumed = 10, and when a is between 10 and 20, then n may be taken = 20. Thus, if 17 is to be subtracted from 63, the remainder, 46, is found by adding 3 ( = 20 — 17) to 63, and rejecting 20 from the sum.

Hence, in trigonometrical calculation, instead of subtracting the log. of the first of four proportionals from the sum of the logs, of the second and third in order to obtain the log. of the fourth term, we may add into one sum the logs, of the second and third terms, and the difference between the log. of the first and 10, and rejecting, mentally, 10 from the sum, we have what is required.

The difference between a logarithm and 10·00000, &c. is called the arithmetical complement of that logarithm; and it is most readily obtained by subtracting every figure, beginning at the left hand, from 9, and the last significant figure from 10; thus, the arithmetical complement of 2·7365) is 7∙26349, and the arithmetical complement of 9∙78460 is 0·21540. By the introduction of the arithmetical complements of logarithms, fewer figures are required in calculations; for, with a little practice, it will be found as easy to read the arithmetical complements of logarithms from the table as the logarithms themselves. We have now this rule for finding a fourth proportional to three given numbers. Add together the arithmetical complement of the logarithm of the first term, and the logarithms of the second and third terms; the sum is the logarithm of the fourth.

***As an example, take the above proportion:

Sin. (B — C): sin. ½ (B + C) = b — c : a.

Sin. ⅛ (B — C) arith. comp, 0-b8465 . , .

Sin. ⅜ (B + C) 9∙78445

b — c................... 2*55630

a = 1680-4 3∙22540

We have the same value of a as above.
[7:21:377]

Cask IV. Given J side b = 5β2 - Remnred the angles

(»idee = 320 J A ’ B ’ C ·

2)1682 Logs.

s = i (a -f- b + c) = 841 7O7Û20 Ar. comp. log. s — a — 41 1’61278 s —0 = 279 2∙44560 s— c = 521 2∙7l684

2)3∙85042

Μ 1∙92521

R 1000000

RM 11∙92521

— _ taii< , a 10 , 3 j2431 A _ 64 o 1 , 55 ,

RM

-^=rtan.⅛B 9∙17961⅜B= 16 47 23 j^t = tan.⅜C 9-20837⅜C = 9 10 41

Verification 89 59 59.

Hence the angles come out

A = 128° 8' 50'', B = 33° 34’ 46", C = 18° 21' 22''. The verification shows the correctness of the result.

Various applications of trigonometry have been given in Mensuration and Navigation; to these we refer the reader for more examples in the calculation of triangles.

Part II.— Spherical Τrigονομετrυ.

1. In our treatise on Geometry we have defined a pyramid (Sect. xi. Der. 6). The most simple solid of this kind is the triangular pyramid, or tetraedron, contained by four plane triangles, or faces, each of which has a side common with the other three. The angles of these form four solid angles. Each of the solid angles presents for our consideration six plane angles; namely, the three angles about the solid angle, and the three angles made by the planes of the faces, which are so related, that any three of them being given, the other three may be found. The investigation of formulae which furnish rules by which this may be accomplished, forms the object of what is to follow.

2. In the calculus of sines, the things considered are plane angles. If two of these be inward angles of a triangle, and the third the outward and opposite angle, our formula for the sines, tangents, &c. of the sum and difference of two angles (Algebra) will express relations among the angles of a plane triangle. We are now to investigate formulae analogous to these, which shall involve the angles of a pyramid. Although the calculus which treats of angles made by straight lines in a plane may be established independently of a circle, yet it is conducive to perspicuity to employ arcs of circles as the measures of angles. This has led to the measuring of the angles of a pyramid by arcs of circles on the surface of a sphere whose centre is the vertex; then the angles made by the planes of its faces will be shown by the angles contained by tangents to the circles at their points of intersection.

3. In treating of spherical trigonometry, we shall begin by explaining some properties of the sphere which depend on principles purely geometrical.

Definitions.

I. A sphere is a solid figure generated by the revolution of a semicircle about a diameter which remains unmoved.

II. The diameter about which the semicircle turns, is the axis of the sphere; and the point which is the middle of that diameter is the centre of the sphere.

III. A straight line drawn from the centre to any point in the surface of a sphere, is a radius of the sphere. It is evident that all the radii are equal.

Theorem I . Every section of a sphere made by a plane is a circle.

Let ADBE be a sphere, and DHE the curve line which is the common section of its surface and any plane. From O, the centre of the sphere, draw OF perpendicular to the plane DHE, and draw FH to any point in the curve DHE and join OH. In the triangle OFH, the angle OFH is a right angle; therefore FH = ^∕(OH- — OF’). Now the lines OH and OF are invariable for all points in the curve DIIE, therefore FH is invariable, and the curve DHE is the circumference of a circle whose radius is FH.

Der. IV. The common section of a sphere and any plane that passes through its centre, is a great circle of the sphere. Let ADBE be a sphere whose centre is O; the circle DFE, in which any plane passing through O meets the sphere, is a great circle.

Cor. 1. All great circles of the sphere are equal; and any two great circles mutually bisect each other.

Cor. 2. A great circle may pass through any two assigned points on a sphere, which are not the extremities of a diameter. For if straight lines be drawn from the centre to the points, a plane may pass along these lines, and only one plane, whose common section with the sphere will be a great circle.

Der. V. The pole of a great circle is a point on the surface of the sphere, from which all straight lines drawn to the circumference of the circle are equal.

Let DEE be a great circle, and O its centre; take D and F any points in its circumference, join OD, OF, and draw OA perpendicular to the plane of the circle, meeting the surface of the sphere in A, and join AD, AF; because AO is perpendicular to the plane DEE, the angles AOD, AOF are right angles; therefore AD = AF, hence A is the pole of the circle DFE.

Cor. Since a perpendicular to the plane of the circle DFE through its centre must meet the sphere in two points; a great circle has two poles, A, B, which are the extremities of a diameter of the sphere perpendicular to the plane of the circle, and it can have no more.

Theorem II. The arcs of a great circle between the poles and the circumference of another great circle are quadi ants.

Let A and B be the poles of a great circle DFE, whose centre is O; let ADB be any other great circle that passes through A and B and meets the circle DFE in E: Because the poles of the circle DEF are at the extremities of AB, the diameter of the sphere that is perpendicular to its plane, the angles AOD, BOD are right angles, therefore AD, BD, the arcs of the circle ADB, arc quadrants.

Der. VI. A spherical angle is the angle on the surface [7:21:378]of the sphere made by two great circles at their intersection: it is identical with the angle made by two straight lines in the planes of the circles which touch them at their intersection; and it is also the same as the angle made by the planes of the circles.

Let AEB, AFB, be two great circles that intersect each other at A and B (Fig. 2). From O the centre of the sphere, and A the intersection of the circles, draw OE and AP in the plane of the circle AEB, and OF, AQ in the plane of the circle AFB perpendicular to AB, the common section of the circles; the straight lines AP, AQ, will touch the circles at A, and form an angle PAQ, which is the spherical angle at A; and since PAQ is equal to EOF (Geometry, Part ii. 10, 1), the inclination of the planes of the circles, this last also expresses the spherical angle at A.

Cor. 1. the adjacent spherical angles which one great circle makes with another, are together equal to two right angles; and the vertical or opposite angles made by two great circles which cut one another are equal.

C or . 2. The spherical angle made by two great circles is measured by an arc of a great circle whose pole is at their intersection.

Der. VII. A spherical triangle is a figure on the surface of the sphere comprehended by three arcs of great circles, each of which is less than a semicircle.

If planes be supposed to pass along the arcs AB, BC, AC, which are the sides of the triangle; these will pass through O, the centre of the sphere, and form at that point a solid angle O (Geometry , Der. vi. of Part ii).

Τηεοrεμ III. Any two sides of a spherical triangle are together greater than the third; and the sum of the arcs which are its sides is less than the circumference of a great circle.

The arcs AB, AC, BC, arc the measures of the plane angles AOB, AOC, BOC, about the solid angle O. Now, any two of these plane angles are greater than the third (Geometry, Part ii. 16, 1), therefore any two sides of the spherical triangle ABC is greater than the third: and because the sum of the plane angles about the solid angle is less than four right angles (17); therefore the sum of the sides of the spherical triangle, which are their measures, must be less than the sum of four quadrants, that is, less than the circumference of a great circle.

Theorem IV. If about the angular points of a spherical triangle, as poles, there be described three great circles on the sphere; these by their intersections will form a triangle which is said to be supplemental to the former; and the two triangles are such, that the sides of the one are the supplements of the arcs which measure the angles of the other.

Let ABC be a spherical triangle; let EF be an arc of a great circle whose pole is A, and FD an arc of a great circle whose pole is B, and DE an arc of a great circle whose pole is C; these three arcs forming the triangle DEF. Let DB be an arc of a great circle passing through the points D, B, and DC an arc of a great circle passing through the points D, C. Because B is the pole of the arc DF, the arc BD is a quadrant (Th . II.), and because C is the pole of the arc DE, the arc CD is a quadrant. Since then the arcs BD, CD, are both quadrants, D is the pole of a great circle that passes through the points B, C; but only one great circle can pass through two points on the sphere not directly opposite, therefore D must be the pole of the arc BC. In the same way, it will appear that E is the pole of the arc AC, and F the pole of the arc AB. Produce the arcs AB, AC, if necessary, till they meet the arc EF in G and H; the point A being the pole of the arc GH, the angle A will be measured by GH. Now EH and FG being both quadrants, EH + FG = 180°; but EH + FG = EF + GH, therefore EF + GH = 180°, that is, EF, a side of the triangle DEF, is the supplement of GH, the measure of the angle A. In the very same way, it will appear that DF is the supplement of the angle B, and DE the supplement of the angle C; and because the triangle ABC may be formed from the triangle DEF exactly in the same way that DEF is constructed from ABC, namely, by describing great circles about the angles D, E, F, as poles, the triangles and their relations to each other may be interchanged. Hence it appears that BC, AC, AB, the sides of the triangle ABC, are the supplements of D, E, F, the angles of the triangle DEF.

Theorem V. The three angles of any spherical triangle arc greater than two right angles, but less than six.

In the triangle ABC (fig. 4), the measures of the angles A, B, C, together with DE, EF, DF, the sides of the supplemental triangle DEF, make six right angles. (Τη. IV.) Now the sum of the sides of the triangle DEF is less than four right angles (Tn. III.), therefore the sum of the three angles A, B, C, is greater than two right angles; and because the three inward angles of any triangle, together with the adjacent exterior angles, are equal to six right angles, therefore the inward angles alone are less than six right angles.

4. We might proceed in this way, establishing the whole doctrine of spherical trigonometry by a series of synthetical demonstrations; such was the practice of the earlier writers on this subject. This however has now given way to a more convenient and elegant method of proceeding, which dispenses with complicated diagrams representing on a plane surface circles on a sphere. It deduces by the angular calculus the whole theory of spherical trigonometry (including also the theorems already here demonstrated) from a single proposition, in a manner probably first followed out by Bertrand of Geneva in his Développement nouveau de lu Partie Elémentaire des Mathématiques (Geneva, 1778). Lagrange seems to have overlooked this work, for in his elegant memoir, entitled Solutions de quelques Problèmes relatifs aux Triangles Sphériques. avec une Analyse complète de ces Triangles, given in vol. ii. of Journal de l’Ecole Polytechnique, he has attributed the theorem to De Gua, who also gave it in the Memoirs of the Academy of Sciences for 1783.

5. Our article on the Arithmetic of Sines (Algebra, sect 25) is an example of what may be done by the power of anaiysis alone in deducing an extensive theory from a few simple geometrical principles. Such is the way in which Lagrange and later writers treat of spherical trigonometry. We shall also follow the method of Delambre, who, in his Astronomie Théorique et Pratique, chap. x., instead of numerous rules deduced from geometrical constructions, for determining when angles or arcs are less or greater than quadrants, considers their sines and cosines as positive or negative quantities, according to the principles of analytic geometry, as explained at article 225 in Algebra. From what has been there explained, it appears that,

(1.) A sine is positive when the arc is less than 180°, but negative when the arc i« between 180° and 360°.
[7:21:379]

(2.) If an arc be taken negatively, its sine is negative in the first half of the circle, but positive in the second.

(3.) The cosine of an arc = 0 is unity: it is 0 when the arc = 90°; between arc = 0 and arc = 90°, the cosine is positive, and between arc = 90° and arc = 270° the cosine is negative, and between 270° and 360° it is positive. A positive and a negative arc have the same cosine, so that to a given cosine the arc may be taken either as positive or as negative.

(4.) Since tan. A≡ the tangent is positive in the first quadrant of the circle, because the sine and cosine are both positive; it is negative in the second, because the sine is positive and the cosine negative; it is positive in the third quadrant, because the sine and cosine are both negative; lastly, it is negative in the fourth quadrant, because the sine is negative and the cosine positive. The tangent is = 0 when the arc is 0 or 180°, and infinite when the arc is 90° or 270°.

(5.) When the arc is known, we can always give the tangent the proper sign. If the arc X is not known, and we have simply tan. x = a, it cannot be known whether the are be positive or negative; if a be a positive number, the tangent belongs alike to two arcs, x and 180° + x. If a be negative, tan. a belongs to an arc between 90° and 180°, or between 270° and 360°. The problem therefore admits of two solutions.

***But if tan. x = -, and that the two terms of the fraction n are positive, then x ∙κ 90°. If m be positive and n negative, x ≥.. 90 β ; when anand n are both negative, x ≥*, 180°; lastly, if m be negative and n positive, x 270°.

(6.) The rule of the signs for cotangents is the same as for the tangents, since tan. x = ——, or cot. x = —-—. cot. x tan.«

(7.) The rule for the signs of secants is the same as that for cosines, seeing that sec. x=—-—.

COS. X

(8.) The rule for cosecants is the same as for sines, since 1

cosec. X = - sin. x

(9.) When an arc is known, the sign of its sine is known; but when x, an unknown arc, is to be found from sin. x — a, the sine belongs alike to x and 180° — x, if a be positive; or to 180° + x, and 360° — x, if a be negative.

(10.) When an arc a is known, the sign of its cosine is known; but if we have cos. x = b, we do not know whether x be in the first or fourth quadrant if x is positive, or whether it be in the second or third when x is negative. The problem therefore has also, in this case, two solutions. The nature of the problem may however determine that which the algebraic rule leaves ambiguous. Delambre, in his Astronomy, says, “These rules are general, and easily remembered, and never mislead a calculator; while, on the contrary’, the rules with which most writers have complicated trigonometry, are so perplexing that there is hardly an author that has not given false ones.” The rule of the signs was long neglected by astronomers, on the known principle that improvements come slowly into general use.

6. *****Let ABC be a spherical triangle, and let O be the centre of the sphere; draw the radii OA, OB, OC. The solid angle at O is contained by the three plane angles AOB, BOC, AOC, and the measures are the arcs AB, BC, AC, which are the sides of the triangle ABC.

At the point A in the plane AOB, draw AD touching the arc AB, and meeting the plane BOC in D; and in the plane AOC draw AE touching the arc AC, and meeting the plane BOC in E. The angle DAE contained by the tangents AD, AE, is the same as the spherical angle BAC (Der. vi.). Join the points D, E by a straight line, which will be at once the base of the plane triangles DAE, DOE, the vertices of which are joined by the line AO, a radius of the sphere. We shall throughout make this line the unit, and express it by 1. Let us now denote the sides and angles of the spherical triangle ABC as follows.

The sides BC = a, AC — b, AB = c; The angles BAC = A, ABC = B, ACB = C. We have now

AE = tan. b — -- p OE = see. b = ., cos. b cos. o

. τ∙x s ∙ n ∙ c AΓ, 1

AD = tan. c — , OD = sec. c = .

cos. c cos. c

By plane trigonometry (Tn. VIII.), in the plane triangle ADE, DE 2 f= AE 2 + AD 2 — 2 AE∙AD cos. A, ( = tan. 2 b 4- tan. 2 c — 2 tan. b tan. c cos. A; and in the triangle ODE, we have also de , f = OE- + OD 1 — 2 OE∙OD cos. α,

I = sec. 8 b -f-sec. 2 c — 2 sec. b sec. c cos. a. Hence, subtracting the sides of the first of these equations from the sides of the second, and considering that sec. 2 b— tan. 2 b = 1, sec. 8 c — tan. 2 c — 1, we have 0=l-∣-l — 2 sec. b sec. c cos. a + 2 tan. b tan. c cos. A; and hence, again dividing the terms by 2, &c.

sec. b sec. c cos. a = I -f-tan. b tan. c cos. A;

, . cos. a , 1 sin. b sin. c , that is,; =1-1 r cos. A; cos. b cos. c cos. o cos. c and multiplying both sides by cos. b cos. c, cos. a = cos. b cos. c -∣- sin. b sin. c cos. A.

This is the formula given by Bertrand, and it comprehends in itself the whole principles of spherical trigonometry.

7. Since the same property must belong alike to the three sides, we may interchange the letters which represent them and the angles, and thus have

I.

Cos. a = cos. b cos. c + sin. b sin. c cos. A...(D Cos. b = cos. a cos. c 4- sin. a sin. c cos. B...(2) Cos. c = cos. a cos. b + sin. a sin. b cos. C...(3)

8. To deduce other formulae from these, w∙e add and subtract the first and second, and thus get

, , f (cos. b + cos. α) cos. c cos. > -J-cos. a — i + ζ s i n , α cos , g s i n . ⅛ c0 s. A) sin. c,

, f — (cos. b — cos. a) cos. c cos. b cos. a — I (s ; n , a cos . β — sin. b cos. A) sin. c∙, and from these again, by transposing the first term on the right-hand side,

, . , , ,. , l’ sin. a cos. B I .

(cos. b + cos. α) (1 - cos. c) = ∣ gin b c θ g a j sin. e, , i ∖ . ∖ I sin. “cos. B I . _.

(cos. b — cos. a) ( 1 + cos.c) = ∣ _ gin b cog a ; sin. c, and taking the products of the corresponding sides, and considering that 1 —cos? c = sin. 1 c, we have, after dividing by sin? c, cos? b — cos. 2 a = sin. 2 a cos.’ B — sin.- b cos? A.

Now, since cos. 2 b -fsin. 2 b = cos. 2 a -fsin. 1 a, we have cos. 1 b — cos. 1 a = sin. s α — sin? b; therefore, sin. s a — sin? b =sin. 2 a cos. 1 B — sin.*' b cos. 1 A; [7:21:380]and sin* a (1 —cos. t B) ≡ sin. t Z> (1 —cos.' A); that is, sin. s a sin.' B = sin.* b sin. 2 A, and sin. a sin. B = sin. b sin. A.

From this last, we form these three formulae,

II.

Sin. Z>_sin. B

Sin. a ~ sin. A ' )

Sin. c_sin. C .

Sin. a sin. A '

Sin. b sin. B

Sin. c - sin. C '

9. By No. 3 of formulae (I.)

cos. c — cos. a cos. b -p sin. α sin. b cos. C, therefore cos. b cos. c = cos. a cos? b -f-

sin. α cos. b sin. b cos. C.

But by No. 1 of formulae I.

cos. b cos c = cos. β—sin. b sin. c cos. A; therefore cos. a— sin. b sin. c cos. A = cos. a cos.’ b -f-

sin. a cos. b sin. b cos. C; and cos. a (1 — cos? ⅛) = sin. b (sin. c cos. A + sin. a cos. b cos. C).

In this expression put sin. 8 b for I—cos. 2 b, and divide both sides by sin. b ; the result is cos. a sin. b z= sin. c cos. A 4- sin. a cos. b cos. C....(wι)

But by formula 2 of II. sin. c ≡ s .- ' r sin. C; therefore sin. A

, . _ cos. A . , cos. a sin. b = sin. α sin. C .-— i ∙ 4- sin. a cos. b cos. C. sin. A

Divide now by sin. α, and put cot. α for and cot. A for sin. α

-T—r, and there is obtained sin. A

cot. a sin. b ∑= cot. A sin. C ψ cos. b cos. C. By changing b into e and C into B, we have also cot. a sin. c = cot. A sin. B -f-cos. c cos. B.

By treating these two formulae like those in articles 7 and 8, we obtain a class of six formulae, viz.

III.

Cot. a sin. b = cot. A sin. C -f-cos. b cos. C (1)

Cot. a sin. c= cot. A sin. B -j-cos. c cos. B (2)

Cot. b sin. a = cot. B sin. C + cos. a cos. C (3)

Cot. b sin. c — cot. B sin. A -∣- cos. c cos. A (4)

Cot. c sin. a — cot. C sin. B -f-cos. a cos. B (5)

Cot. c sin. b — cot. C sin. A -f-cos. b cos. A (6)

In these formulae, the letters which represent the parts of the triangle are not so symmetrically arranged, as in the two former groups; and on that account they are less easily remembered. This inconvenience will be obviated by attention to the following remarks.

(1.) Both members of each formula begin with the product of a cotangent and a sine.

(2.) The first member is formed from any two of the three sides, and the first term of the second member contains an angle opposite to one of these, and the angle opposite to the side which is not in the first member.

(3.) The last term of the second member is formed by the two cosines of the same arcs of which the sines are already in the equation.

10. Resuming equation (m ) in the preceding article, viz. cos. α sin. b ~ sin. a cos. b cos. C + sin. c cos. A, change α into b and b into α, also A into B, and we have sin. a cos. b = sin. b cos. α cos. C -fsin c cos. H; and multiplingboth sides by cos. C, sin.α cos. b cos. C = sin. b cos. a cos. 1 C -∣- sin. c cos. B cos-C. This value of sin. a cos. b cos. C being substituted in the former equation, it becomes cos. α sin. b = sin. b cos. a cos. , C -fsin. c (cob. B cos. C + cos. A), and by transposing the first term of the second side, cos. α sin. b ( I —cos? C) = sin. c (cos. B cos. C -f-cos. A ). that is, cos. a sin. A sin. 2 C = sin. c (cos. B cos. C -(- cos. A).

Now by (3) of formulae IL, sin. b sin. C = sin. c sin. B, therefore this last expression being substituted in the first side of the equation instead of its equal, and both sides being divided by sin. c, we have cos. a sin. B sin. C = cos. B cos. C -f-cos. A.

And by treating the expression just found, as we did those in articles 7, 8, 9, we have these formulae:

IV.

Cos. A = —cos. B cos. C -Jsin. B sin. C cos. a (1)

Cos. B = — cos. A cos. C + sin. A sin. C cos. b...... (2)

Cos. C = — cos. A cos. B -fsin. A sin. B cos. c (3)

11. The sides of a spherical triangle being still denoted by a, b, c, and its angles by A, B, C, let A', B', C', and a', lit, d, be arcs or angles, such that a + A' = 180°, b + B'= 180°, c⅛C'= 180°; A + α' = 180°, B + b' = 180°, C + d = 180°.

Then we have cos. a = — cos. A', cos. b — — cos. B', cos. c = — cos. C', sin. α — sin. A', sin. b = sin. B', sin. c = sin. C'; cos. A — — cos. α', cos. B = — cos. ∕∕, cos. C = — cos. d, sin. A = sin. a', sin. B = sin. fr, sin. C — sin. c.'

Let these values of the sines and cosines of a, b, c and A, B, C be substituted in formulae I. and IV.; they then become cos. A' = — cos. B' cos. C' + sin. B' sin. C' cos. α', cos. B' = — cos. A ’ cos. C' -Jsin. A' sin. C' cos. b', cos. C' = —cos. A' cos. B' 4- sin. A' sin. B' cos. d; cos. a' — cos. U cos. d -∣- sin. b' sin. d cos. A', cos. b’ — cos. α ’ cos. d -Jsin. a' sin. d cos. B', cos. d — cos. a' cos. b' -fsin. a' sin. b' cos. C'.

Hence, from I. and IV., if the arcs a ', b’, d be taken as the sides of a spherical triangle, then the arcs A', B', C' must necessarily be the measures of its angles; thus we have proved analytically our fourth theorem, which has been demonstrated in article 3, from considerations purely geometrical. This theorem, which has been known for two centuries past, may be expressed thus: Any spherical triangle may be changed into another of which the sides and angles are respectively the supplements of the angles and sides of the first. This is the supplement or polar triangle.

12. The four sets of formulae which have been investigated in articles (6), (7), (8), (9) (10), are sufficient for resolving all cases of spherical triangles. When one of the angles is a right angle, they become more simple. We shall now adapt them to that case.

Right-Angled Spherical Triangles.

Let us suppose that A — 90°, then cos. A and cot. A are each = 0, and sin. A = 1.

From 1 and 2 of form. I. cos. a = cos. b cos. c.

*****l’ . n sin. 6

I sin. B = -—,

From 1 and 2 of II. <(

. _ sin. c sin. C =

sin. α

From 1 and 2 of III. cot “sin ' b = cos ’ k o6ft cot. a sin ∙ = cos. ccos. B.

From 4 and 6 of III. cot ∙ b s " κ l’ = C< *

cot. e sin. b z= cot. C.
[7:21:381]

J cos. B cos. C = sin. B sin. C cos. a, cos. B = sin. C cos. b, ( cos. C = sin. B cos. c.

By considering that the cotangent of an arc is the reciprocal of the tangent, and that the tangent is e<pial to the sine divided by the cosine, we now obtain these ten formulae for right-angled triangles, in which A is the right angle.

V.

Cos. a = cos. b cos. c (1)

Sin. B = ~, sin. C = (2) (3)

sin. a sm. α r , ,, tan. c „ tan. b ... ...

Cos. B = , cos. C = - (4) (5)

tan. a tan. a ' v

π ’ τ> tan. b tan. c . . ea∖ sin. c sin. b

Cos. a = cot. B cot. C (8)

„ , cos. B cos. C

sin.C sm. B

13. These ten formulae, which may however be expressed in words by six, are of continual use in astronomy. It is therefore desirable that they should be comprehended in the fewest number possible of practical rules, simple and easily remembered. Such were found by Napier, the inventor of logarithms, and are known by the name of Napiers Rules of the circular Parts. They are not new propositions, but are merely enunciations, which, by a particular arrangement and classification of the parts of a triangle, include all the six proportions. They are perhaps the happiest example of artificial memory that is known.

In a right-angled spherical triangle ABC, setting aside the right angle A, there remain the sides AC = b and AB = c about the right angle; the hypotenuse BC = a, and the two oblique angles B, C. The first two of these, viz. the sides b, c, and the complements of the remaining three, viz. 90° — a, 90° — B, 90° — C, are called the five circular parts. They are called circular, because, when taken in their order, they go round the triangle.

14. When one of the five circular parts is taken as the middle part, the two next to it, one on each side, are called the adjacent parts; and the other two, each of which is separated from the middle by an adjacent part, are called the opposite parts.

Thus, if 90° — a be the middle part. 90° — C and 90° — B are the adjacent parts, and b and c the opposite parts; if 90° — B be the middle part, 90° — a and c are the adjacent parts, and 90° — C and b are the opposite parts; and if c be the middle part, 90° — B and b are the adjacent parts, and 90°— a and 90° —C the opposite parts; and so on.

This arrangement being made, the rule of the circular parts is contained in the following proposition:

The rectangle under radius and the sine of the middle part is equal to the rectangle under the tangents of the adjacent parts; also to the rectangle under the cosines of the opposite parts.

It may assist the memory if we represent the middle part by the letter M, the adjacent parts by the letters A, a , and the opposite parts by the letters O, o ; the rule will stand thus:

****R∙sin. M = tan. A ∙ tan. a; R ∙ sin. M — cos. O ∙ cos. 0.

15. To verify the rule, let 90° — a, the complement of the hypotenuse, be the middle part; then 90° — C, and 90° — B will be the adjacent, and b, c the opposite parts; and by the rule,

R ∙ sin. (90° — a) f = tan ∙ ( 90 ° — b ) tan ∙ (9υ ° — C)

v 7 I = cos. ο cos. e; ∏ . λλ , λ f = cot. B cot. C. This 19 form. (8) ( = cos. 0 cos. c this is form. (1)

Let 90° — B, the complement of one of the oblique angles, be the middle part, then 90° — α and care the adjacent parts, and 90° — C and b the opposite. The rule gives R ∙ cos B i = c0t ∙ α tan ∙ c Fonn ’ < 4 ’ 5 >

I = sin. C cos. b Form. (9, 10)

Let b, one of the sides about the right angle, be the middle part, then 90*— C and ewill be the adjacent parts, and 90° — a and 90° — B the opposite parts. In this case, by the rule, n . , f = cot. C tan. c.......... Form, f6, 7)

R ∙ sin. bl _ . . n t , )

I = sin. a sin. 15 r orm. (2, 3)

These are all the cases of right-angled triangles.

In applying the rule of the circular parts to resolve any case of right-angled triangles, consider which of the three quantities named (the two things given and the one required) must be made the middle term, in order that the other two may be equidistant from it, that is, may be both adjacent or both opposite; then the one or the other of the theorems which constitute the rule will give the value of the thing required.

Suppose, for example, that c a side about the right angle, and a the hypotenuse, are given to find the angle C. It appears that if c be made the middle part, then 9(90° — a and 90° — C are the opposite parts; and therefore that R ∙ sin. c = sin. a sin. C, and sin. a : sin. c = R: sin. C.

16. Delambre has avowed that he found it easier to remember the six formulae for right-angled triangles than to apply Napier’s rules. In astronomy, the sun’s longitude ( ) is the hypotenuse of a right-angled triangle, of which the right ascension (A R) and declination (D) are the sides about the right angle, the obliquity of the ecliptic (ω ) is one of the oblique angles. Let the other oblique angle be denoted by A, then the six formulae for right-angled spherical triangles adapted to astronomical calculation will stand thus:

***Sin. D = sin. ω sin. Q, tan. .4R = cos. ω tan. 0, tan. D = tan. u sin. All, cos. o = cos. AR cos. D, cos. Q — cot. a, cot. A, cos. A = sin. ω cos. AR.

The practical astronomer, who has continual occasion to use the first four of these formulae, will find no difficulty in retaining them in his memory. But Napier’s rules will probably be most easily remembered by students of that science.

17. The general formulae, I., II., III., IV., serve for all spherical triangles; but they admit of transformations, which make them more convenient for logarithmic calculation.

Resuming the formula, (I.) we put the first under these two forms.
[7:21:382]

Cos.o= (cos.⅛cos. e —sin.6sin.c)4-sin.6sin.c(l -J-cos.A), Cos. a= (cos. b cos. c -)- sin.Z>sin.c)—sin.∕>sin.c(l —cos. A).

Now, by the calculus of sines (Algebra, articles 239 and 248),

Cos. b cos. c — sin. b sin. c = cos. (b -fc),

Cos. b cos. c 4- sin. b sin. c = cos. (b — c),

1 + cos. A — 2 cos. 2 J A, 1 — cos. A = 2 sin. 8 ,∣A.

Hence, by substituting, we get

Cos. a — cos. (b -∣- c) 4- 2sin. b sin. c cos. 2 4 A,

Cos. a — cos. (Z> — c) — 2 sin. b sin. c sin. 2 J A; and transposing, and observing that by the calculus of sines (Algebra, art. 240),

Cos.α — cos. (b + c)= 2 sin.] (a -⅜- A 4. e)sin.∣ (6 -f-c—o), Cos. (A —c) — cos,α= 2 sin.) (α-∣-c— b) sin. ; (a -∣-⅛ — c), we have

Sin.osin. ccos. 2 jA=sin.A(α 4- b -f-c) sin.∣(⅛-f-c—σ), Sin. A sin. c sin. ’ ⅛ A = sin. J (a -Je — b) sin. ; (a + l > — c)∙

To abridge, let us put À (a -f- b -fc) ≡ s; then J (b 4- c — a) = s — a, ι (a -j- c — b) — s — b, J (α + b—c) - s — c. i. , , . sin. s sin. (s—a)

>v c have now cos.’ 4 A = r A:

sin. b sin.c sin.’ ⅛a = ⅛4» ⅛L(i=-2).

sin. b sin. c

There are expressions exactly like these for the squares of the cosines and sines of the halves of the angles B, C; and from the whole we obtain

Cos.JA = ∕ g ⅛*⅞∙i*- a V

V sin.o sin. c r , . _ /sin. s sin. (s — b) z .

Cos.⅛B = ∕ ——⅞ ' h (α )

¼ ʹ sin. asm. c ; ,

Cos.lC = ∕ s A⅛(Lt±) .

* sin. a sin. b

Sin.⅛Λ = ∕ g i n J⅛. ¾ gin ∙(*-*) , ’

sin. o sin. c

Sin.]B = L...(tf)

3 √ sin. α sin. c

Sin.lC= ∕⅛-⅛∙<*- z, >.

z√ sin. α sin. b

18. Because tan.* JA = —therefore i cos. s ⅛A

tan.UA zz sin.(s-⅛)sin.(>- c )

“sin. s sin. (s — a)

1 . sin. (s — a) sin. (s — b) sin. (s — e) = sin. 2 (s-0) »»i. s √f sin.(s—a) sin. (s—⅛)sin.(s—c) l

1 — —8in7Γ 2 - f’ and it is observable that in this expression the sides a, b, c enter exactly alike. We have now tan. ∣A — ~. By putting B and b, also C and c, successively, instead of A and α, we find like expressions for the tangents of ∣B and ⅜C. Thus we have these three formulae:

tan. 1A = -——V tan. ⅛B = -.—-. yτ> I

1 sin.(sα) 2 sm.(f- b) 1

tan. 1C = -—7 ..

sin. (s— c) J

The formulae in this and the preceding article arc exactly like those found for the angles of plane triangles. Those for the tangents of half the angles (which we believe to be given in this form for the first time) seem to be most convenient for use in all cases. In using the others, if half the angle sought be nearly a quadrant, its cosine ought to be sought
rather than its sine; and
the reverse, if the angle be small.

19. We may treat our fourth group of formulae, of which the first is

Cos. A = — cos. B cos. C + sin. B sin. C cos. σ, in all respects as we have the first, putting it under these forms:

Cos. A = — (cos. B cos. C — sin. B sin. C)

— sin. B sin. C (1 — cos. a),

Cos. A = — (cos. B cos. C -∣- sin. B sin. C)

+ sin. B sin. C ( 1 4- cos. a); from which we have, by the calculus of sines (Algebra, art. 239), &c.

Cos. A -∣- cos. (B -(- C) = — sin. B sin. C (1 — cos. a),

Cos. A + cos. (B — C) = sin. B sin. C (1 -f-cos. a). Now

Cos.A 4- cos. (B+C)= 2 cos.,] ( A+B-f-C) cos,⅛(B 4- C—A ), C os. A -∣- cos. ( B—C ) = 2 cos.^( A -fC—B) cos.( A 4- B—C), 1 — cos. a — 2 sin. s Jσ, 1 4- cos. a = 2 cos. 2 ∣a.

Therefore, putting

S=J(A + B + C), so that S— A = ⅜(B-(-C — A),

S —B= (A + C —B),

S — C= J(A + B —C), we have

Sin. B sin. C sin.’ ⅛a = — cos. S cos. (S — A),

Sin. B sin. C cos. 2 ⅛a = cos. (S — B) cos. (S — C).

There are like expressions which involve the squares of the sines and cosines of lb and lc, from all which we deduce

, ∕ — cos. S cos. (S — A)'

s,n ∙≡ α = √ sCT⅛

sin / — cos. S cos. ( S — B) I

V sin. A sin. C

Sin, is = /=.?»· S cos-(S-C)

“V sin. A sin. B

∕cos∙(S—B)cos. (S — C)

Cos. 4 a — ∕ 7 ⅞, .—τ~ »

i ¼∕ sin. B Sin. C

c∙>-i i =√ i2h ⅛⅛≡⅛= 3 ∙ h« Coslc- ∕ co3 ∙( s ~ a ) c o8 ∙( S- b )

’ ' 1 λ < sin. A sin. B ' J

20. Dividing now the cosines of the halves of the sides by their sines, thereby getting expressions for the tangents, and proceeding as in article 18, putting’

∕ (cos.(S—A)cos.(S — B)co$.(S — C)"J

m =√ γ —cos.S i’

we have these formulae:

(∕)

, ∙ M' ... M'

^-θt∙ .jΛ — ∕n A V Cθt∙ .)b — r ,-- ∕Q T3∖ ’ i cos. (S — A) cos. (S — B)

, M' cos.(b — C)

The sum of the three angles of a spherical triangle always exceeds two right angles; therefore S, half their sum, exceeds a quadrant; and hence cos. S will be a negative quantity, and — cos. S will be positive, and its square root a real quantity.

In astronomical calculations, it very seldom happens that the angles are wanted, from the three sides being given.

21. Recurring to the formulae (a), (∕3) of article 17, we deduce from them the following equations.

Sin. ⅛Bcos. |C _ sin, (s — c) _ sin. ¾ (a + b — c)

cos. IA sin.α - sin. α ’

Sin. ^ C cos. ^B sin, (s — ⅛) _ sin, à ( n + c — cos. ⅛A - sin. a sin. a [7:21:383] Cos.’B cos. ^C _sin. s _ sin. ¾ (o -⅜- ⅛ -f-c) sin. JA - sin.α - sin. a ’ Sin. IB sin. i C_ sin. (s — a)_ sin. ⅛ (⅛ + c — o) sin. ^A sin. a sin. a

By adding and subtracting the sides of these equations, and observing that

Sin. AB cos. JC ± sin. JC cos. JB = sin. 1 (B =±z C),

Cos. JB cos. ! 2 C=p sin. ,JB sin. ! 2 C = cos. ) 2 (B ∑±z C), Sin∙i (a + h —c)-∣-sin.J (aψc— b) = 2 sin.,], “cos. l(Z∣—c), Sin∙i (α-f-⅛—c)—sin. J (a-j-c—b) = 2cos∙iasin.⅜(0— c), Sin. J(a-f-Z>-pc) — sin. ,', (b-(-c—a) = 2sin. ∖α cos. ) 2 (i-∣-c), Sin. <5(6+«+a)+sin. J (Z∣-(-c—a) = 2 cos. ' 2 a sin. ⅛ (⅛-f-c), Sin. a = 2 sin. Ja cos. ⅛a,

(as has been proved in Algebra, articles 240 and 247), we obtain the following results:

(*)

Sin. J 2 (β -tC)_ cos. I (⅛ — c) (1 )

cos. Ü - ' cos. Jα

Sin.J(B— C)_sin.|(é — c) cos. JA - sin. ⅜a Cos. j(B -⅜- C)_ cos. ⅜(⅛ + c) . g , sin. IA - cos. ∣a

Cos.⅛(B—C)_ sin. ⅜ (b -f- c) sin. ⅜A - sin. ∣a

These formulae were first given by Gauss, in his work Theoria Motus corporum Caelestium.

22. Let us now divide the sides of formula (1) by those of formula (3), also the sides of (2) by the sides of (4), and the sides of (4) by the sides of (3), and, lastly, the sides of (2) by the sides of (1); and put tan. in the results instead of —also cot. instead of °∙~'; we then have cos. stn∙

(∙)

Tan. j(B + C) _ cos. ⅜ (⅛ — c) ae cot. I A cos. ⅛ (b -p c) Tan.⅜(B-C) _ sin, ⅛(b-c) . g . cot. ⅜A - sin. ∣(6 -fc)

Tan. ⅜(6-⅜- c) _ cos. ⅛(B—C) tan.⅜α ~ cos. ∣(B -∣- C) Tan.⅜(⅛-c)_ sin.⅜(BC )

tan. ⅜α - sin. ⅜ (B -∣- C) ' ‘

These formulae, expressed in words as proportions, are commonly called Napiers analogies. He however only gave the last in the concise form which it here presents. The third, as he left it, had two common divisors in the terms of a ratio, which he overlooked. His commentator Briggs removed the defect, and, in addition, deduced the first and second formulae from the third and fourth by the theorem here demonstrated in art. 11. Thus he is entitled to share with Napier in the honour of having discovered these very elegant properties of spherical triangles.

From the first and second of these formulae, there is found, by division, this other formula:

(<)

Tan. 4(B - fC) _ tan. ∣(⅛ -∣- c) Tan.∣(B — C) - tan. ⅜(Z> — c)'

This formula is also due to Napier.

23. There are some properties of spherical triangles which may be more concisely proved by the formulae than by geometrical reasoning. Such are those which follow.

I. In a spherical triangle, the greater angle is opposite to the greater side, and conversely.

Let α, b, c be the sides of a spherical triangle, and B, C the angles opposite to the sides b, c. It appears from formula (2) of δ , article 21, that

Sin. ⅜(B — C)_ cos.⅜A.

Sin. ⅜(Z∕— c) - cos. ⅛α

The second member of this equation is always a positive quantity, therefore the first must also be positive. This can only be true on the hypothesis that when b is greater than c , then B is also greater than C. The truth of the converse follows from the same principle.

II. the angles at the base of an isosceles triangle are equal, and conversely.

For we have

Cos. ½ a sin. ½ (B — C) = cos. ½ A sin. ½(b — c ). When b = c, the second side of the equation becomes = 0, therefore the first side must also be = 0. This requires that B = C. The converse follows from the same formula.

III. Any two sides of a spherical triangle are together greater than the third.

It appears from article 17, that

****. sin. b sin. c cos. 2 i A

2y ~ j sm.⅜(α -I- b -fc)

The second side of this equation is always a positive quantity, therefore the first must also be positive. This requires that b + c be in every case greater than a. This proposition has been otherwise proved geometrically, Theorem III.

IV. We have found (article 21) that

Cos. ½ (b + c ) sin. ½A = cos. ½ (B + C) cos. ½ a .

Now sin. ½A and cos. ½ a are always positive, because each arc is less than a quadrant. Therefore cos. ½(b + c ) and cos. ½ (B + C) have always the same sign; and hence ½ (b + c) and ½ (B + C) are of the same affection, and B + C, b + c, either both less than a semicircle, or both equal to it, or both greater.

V. The hypotenuse a of a right-angled spherical triangle is less or greater than 90°, according as the sides b, c, about the right angle A, are of the same affection, or of different affection; that is, in the former case, both less or both greater than 90°; and in the latter, one less and the other greater than 90°.

This follows immediately from the formula cos. a ≡ cos. b cos. c (art. 12); for if b and c be both less than 90° or both greater, they have like signs, and their product has the sign +, which indicates that a is less than 90°. If one of the two b, c be less and the other greater than 90°, they will have different signs, and the sign of their product will be —, and a must be greater than 90°.

VI. The same rule applies to each of the sides b, c ; for b will be less than 90° if a and e be of the same affection, but greater than 90° if a and c are of different affection.

VII. The formula cos. a = cot B cot. C shows that the hypotenuse a will be less than 90° when B and C are of the same affection, but greater than 90° if they be of different affection.

VIII. Also, that either of the angles B, C, will be less than 90° if a and the other angle be of the same affection, but greater than 90° if they be of different affection.

IX. The formula cos. B = cos. b sin. c shows that the angle B and the opposite side b are always of the same affection.

X. The formula tan. c = tan. a cot B shows that the hypotenuse a and either side c are of the same affection if the angle they contain be acute, but of different affection if the angle be obtuse.

XI. Lastly, the formula tan. b = sin. c tan. B proves that the side and the opposite angle are always of the same affection.

XII. In any spherical triangle ABC, draw AD an arc of a great circle perpendicular to the base BC, thus forming two right-angled triangles ADC, ADB, which have [7:21:384]a common side AD. We have (art. 13) tan. AD = tan. C sin. CD = tan. B sin. BD; hence the angles ACD, ABD, are either both acute or both obtuse. If therefore ACB and ABC be of the same affection, the perpendicular falls within the triangle ABC; but if the angles ACB' and AB'C are of contrary affection, the perpendicular falls without the triangle AB’C.

Solution of Right-Angled Triangles.

A spherical triangle may have one right angle, or it may have two, or three. If it have two right angles, the sides opposite to these will be quadrants, and the third the measure of the remaining angle. If it have three right angles, it will be equilateral, and its sides quadrants. We shall set aside these cases, and consider only that in which one of the angles is a right angle, and the other two oblique, which may be either acute or obtuse. The side opposite to the right angle is named the hypotenuse.

The solutions are to be derived from the formulae in art. 12; and as in the investigations, to abridge, the radius was supposed to be the unit, we shall in the rules express it as in the formulae of plane trigonometry by the letter R.

Case I. Given the hypotenuse a and b , one of the sides about the right angle, to find the other side c, and the oblique angles B, C. These are to be found from formulae (1), (2, 3), (4, 5).

Solution.

***_ cos. α . sin. b tan. b .

Cos. c = r R, sin. B = R, cos. C ≡ IL

cos. b sin. α tan. n

If a and b be of the same affection, then c must be less than a quadrant, and also the measure of the angle C. But if a and b be of different affection, then the side c and the angle c must be both greater than quadrants. The angle B and the side b will be of the same affection.

Case II. Given the sides b and c about the right angle, to find the hypotenuse a , and the angles B, C. Here the formulae to be used are (1), (6, 7).

Solution.

,, cos. ύ cos. c r , tan. 6 „ _ tan. c „

Cos. a= „ , tan. B=- R, tan. C = . . R.

R sin. c sin. o

In all the cases it must be considered whether the arcs found be less or greater than quadrants, by the algebraic signs + or -, which belong to the given arcs and angles as expressed in article 5.

Case III. Given the hypotenuse a, and an oblique angle B, to find the sides b, r, and the other oblique angle C. The formulae to be employed are (2, 3), (4, 5 ). Solution .

***, sin. a sin. B tan. fl cos. B

Sin. b = ∩ , tan. c = ∩ , cos. Q _ tall ∙ j; t tan. a Case IV. Given a side b and the opposite angle B, to find the hypotenuse a, the side c , and the anglc C. This is resolved by (2, 3), (6, 7), (9, 10).

Solution.

... sin. b , 1 . tan. b r , . _ cos. B _

Sin. a = -—∏ R, sm. c = ≈ R, sm. C ≡ — R.

sin. B tan. B cos. b

In this case each of the things required has two values, because an arc and its supplement have the same sine. Hence two distinct triangles may be found from the same data; this is called the ambiguous case of right-angled spherical triangles.

Case V. Given a side b with the adjacent angle C, to find the hypotenuse a , the side c, and the angle B. These are to be found by (4, 5), (6, 7), (9, 10).

SolutioN.

****_ tan. b n sin. b tan. C

I an. a ~ — K, tan. c = , cos. C R

τ , cos. b sin. C cos. B = 5 ——.

it Case VI. Given the two oblique angles B, C, to find the hypotenuse a, and the sides b, c. These are found by (8), (9, 10).

SolutION. r , cot. B cot. C . cos. B τ , cos. C „

Cos. a = 1 ,—-, cos. b = ——-, R, cos. c — 5 R.

R sm. C sin. B

In addition to what has been said in regard to the arcs or angles sought being less or greater than quadrants, it may be useful to remember, that when the thing to be found by the formulae is either a tangent or a cosine, and when of the tangents or cosines employed in its computation, one only is an obtuse angle, the angle required is also obtuse. In what follows we shall consider R = 1.

Oblique-Angled Spherical Triangles.

24. Before entering on these, we shall resolve this problem.

PROBLEM. Let M, N, P be given quantities, and x an unknown arc or angle; it is proposed to find x from this equation:

M cos. x + N sin. x = P.

****Let us assume M = v cos. φ, N ≡t v sin. p; where v denotes a number, and φ an angle, both to be determined. Our equation may now be expressed thus.

v (cos. φ cos. x -fsin. <p sin. x) = P; p hence v cos. (x — φ) — P, and cos. (x — p) = -.

XT . t>sin. φ N ...

r vcos.φ M v , an; ν _ _M N_

- cos. φ~ sin. φ P P

Therefore cos. (tr∙—p) = — cos. φ = — sin. p....(2)

The angle φ is determined by the first equation, and the angle (x — φ) by the second; therefore x = (x — φ) -f∙ p is found.

If we assume M = v sin. φ, and N = “cos. φ. we have t> (sin. φ cos. x -j-cos. p sin. x) — P, and v sin. (x -f- φ) — P; then cot. p= (1)

P p and sin. (x -j- φ) = ~ sin. φ — — cos. φ (2)

Thus φ and x — (x -∣- φ) — φ are both determined.

The assumptions tan. φ or cot. φ are evidently always possible, since an arc φ may be found whose tangent or cotangent shall have any real value whatever.
[7:21:385]

25. There are six cases of oblique-angled triangles, which may be resolved by the formulae which have been investigated; but in some cases they will be modified so as to adapt them to logarithmic calculation.

Case I. Given the three sides a, b, c, to find the angles A, B, C.

Solution. *******Find s = ⅜ (α -∣- b -(- e), and M = ∕-[ sin ∙ <* ~ ≡⅞⅛.- ft λ gin ∙ <' ~s) I; then *r ( Sin. 5 J

Tan. ⅜A = * tan. ⅜B = .∙∙

a sin. (5 — α) - sin. ($ — b)

Tan.∣C= ςr -£ v

• sin.(r — c)

This solution has been found in article 18.

Case II. Given two sides a and b, and the angle A opposite.

(1.) The angle B may be found from formula II. thus:

_. τ , sin. Asin. b Sm. B= i sin. «

The affection of B is ambiguous, unless it can be determined by the rule A, 4- B is greater or less than 180°, according as a -∣- b is greater or less than 180° (art. 23).

(2.) The side c is to be found from this equation: [cos. b ] cos. e -f- [sin. b cos. A] sin. c — cos. a.

Comparing this with the equation of art. 24, viz. M cos. x -fN sin. x = P, we have

M = cos. b, N = sin. b cos. A, P = cos. a; and following the analysis by which the angle x has been there determined, we find that if

Tan. p = tan. b cos. A,

, . . cos. α cos. a cos. a sin. ®

then cos. (c — φ)= . r = . -. ζ.

' cos.o sin. o cos. A

The first value of cos. (c— φ ) being the more simple, is to be used. The arcs c — φ and φ being now both known, c is known.

*****3. To find the angle C, we have, from formula III., [cos. ⅛] cos. C -∣- [cot. A] sin. C = cot. a sin. b.

Here, referring to article 24, we have M = cos. b, N=cot. A, P ≡ cot. a sin. b ; therefore, assuming cot. A Ian.©— z , cos. b we have

„ . , cot. a sin. b sin. a

Cos. (C — φ) = cot. a tan. b cos. f — -^∙ ∙

From either of these formulae the angle C is actermined.

Case III. Given the sides a, b, with the included angle 'C, to find the angles A, B, and side c.

(1.) The angles A, B, may be found from two of Napier’s analogies in article 22. Thu3 we have

T∙"l<A+≡) = oo∙-iC

τ *"∙l(∙ i -I>) = oo∙'iC⅛4⅛⅛J∙

Having now half the sum and half the difference of the angles A and B. each may be found by a known process. They may be otherwise found, by subsidiary angles, from formula I.

(2.) The side c is to be found from the equation [cos. a] cos. b -f- [sin. a cos. C] sin. 5 = cos. c.

Here we have M = cos. a, N = sin. a cos. C, P = cos. c; we therefore make tan. φ — tan. a cos. C.

, . cos. c cos. φ

Then cos. (b — φ) =;

' r cos. α

, , cos. a cos. lb — a) and hence cos. c — -'r cos. f Case IV. Two angles A and B, and c the side between them, are given, to find the sides a, b, and angle C.

(1.) The sides a, b may be found from Napier’s formulae, article 22, thus,

Tan. J(α + fc) = tan. i c∙

Tan. ⅜ (a _ b) = tan. ⅜ c g∙

Half t)ιc sum and half the difference of the sides a, b being found, each is thereby determined.

(2.) The angle C is found from the formula

[sin. A cos. c] sin. B — [cos. A] cos. B = cos. C. Here M = sin. A cos. c, N = cos. A, P = cos. C. We now assume

Cot. φ ≡ tan. A cos. c.

.,∙ι /n , cos. C sin. a

1 hen sin. (B — 0) = r— r -,

' r ' cos. A

, „ cos. A sin. (B — φ) and cos. C = ι — i sin. φ Case V. Two angles A, B and a side a opposite to one of them, are given, to find the sides b, c, and the angle C∙

(1.) The side b is found by this formula:

, sin. a sin. B Sin. b = — 1 —. sin. A

(2.) The side c is found by this formula:

[cot. π] sin. c— [cos. B] cos. c = cot. A sin. B.

Here M — cot. a, N = cos. B, P = cot. A sin. B; we therefore assume f , . cot. a Cot. ffi = e cos. B’

and have sin. (c— φ) — cot A tan. B sin. f.

(3.) The angle C is found from the formula

[sin. B cos. λ] sin. C — [cos. B] cos. C = cos. A. Here M = sin. B cos. a, N ≡ cos. B, P = cos. A.

We assume cot. φ = tan. B cos. a II., . , . cos. A sin. β

lhen sin. (c — C) = κ— c .

' r ' cos. B

Case VI. Given the three angles A, B, C, to find the sides a, b, c.

The solution of this case has been found in art. 19.

Find S = J (A + B + C), and M ’ = ∕∫ co8 ∙ (S — A) cos. ( S — B) cos.(S-C) ^> * ʹ I cos. S J ’

M ’ M'

Then cot. ⅛ a = zg r r, cot. ⅜ b = - ε nλ , ’ cos. (S — A) * (cos. b — B)

t . _ M'

* CO3. (S C)

These formulae arc altogether analogous to those by which the angles are found from the sides. This case rarely occurs in the application of spherics to astronomy.

The analytical artifice of subsidiary angles, by which the formulae are adapted to logarithmic calculation, corresponds to the resolving of an oblique spherical triangle into two right-angled triangles, by an arc of a great circle drawn from its vertex perpendicular to its base.

For the application of spherical trigonometry, see Practical Astronomy, chap. i. (vol. iv. page 63).

The treatises on trigonometry are innumerable. The most considerable is that of Cagnoli, in Italian, of which there is a French translation. On the history of trigonometry, and the progress of its improvement, the reader will find the most ample information in Delambre, Histoire de l’Astronomie Ancienne, Histoire de l’Astronomie du Moyen Age, and Histoire de l’Astronomie Moderne. See also his Astronomie Théorique et Pratique, vol. i. chap. 10.