THE ELEMENTS

OF

SPHERICAL TRIGONOMETRY.

BY JAMES HANN, J,\TE MATHEMATICAL MASTER OF KING'S COLLEGE 8CHOOL, LONDON.

gi |Ufo ^bition, mrb Corwcttb, BY CHAELES H. BOWLING, C.E., LATE OF TBINITY COLLEGE, DUBLIN.

LONDON: VIRTUE BROTHERS & CO., 26, IVY LANE, v ' ' PATEItNOSTEIt HOW. 18GG.

PREFACE TO THE

SECOND EDITION.

The first edition of Mr. Hann's " Elements of Spherical Trigonometry" has been carefully revised, and corrected, and several practical problems from the prize questions of different universities have been added to the work. The author states, in his preface to the former edition:— "In the compilation of this work, the most esteemed writers, both English and foreign, have been consulted, but those most used are De "Fourcy and Legendre. "Napier's ' Circular Parts' have been treated in a manner somewhat different to most modern .writers. The terms conjunct and adjunct, used by- Kelly and others, are here retained, as they appear to be more conformable to the practical views of Napier himself." This edition is recommended with increased confidence to the judgment of all mathematical teachers and students, especially naval instructors and students of naval colleges.

0. H. DOWLINGr, C.E. p

CONTENTS.

Pag© Definitions 1 Polar Triangle 2 Fundamental Formula 4 Relations between the Sides and Angles of Spherical Triangles . . 5, 7 Napier's Analogies 8 Right-angled Triangles 10 Napier's Circular Parts 11 Solution of Oblique-angled Spherical Triangles 18 Ambiguous Cases of Spherical Triangles . . . ' 23 Table of Results from the Ambiguous Cases 25 To reduce an Angle to the Horizon 26 Numerical Solution of Right-angled Spherical Triangles 28 Quadrantal Triangles 38 Oblique-angled Triangles 35 Area of a Spherical Triangle . 44, 48 Girard's Theorem 51 Legendre's Theorem 52, 55 Solidity of a Parallelepiped 55 LexelFs Theorem 57, 59 Polyhedrons 61, 67 Examples and Problems 67 SPHEEIGS. r.

PRELIMINARY CHAPTER. ]. A sphere is a solid determined by a surface of which all the points are equally distant from an interior point, which is called the centre of the sphere. 2. Every section of a sphere made by a plane cutting it, is the arc of a circle. c—— Let 0 be the centre of the sphere, A P B A a section made by a plane passing through it, draw 1 00 to the cutting plane, and pro¬ duce it both ways to D and E, and V N M draw the radii of the sphere 0 A, OP. ' Now, since OOP and OCA are right anales, O A2 — 0 C2 = A C2. and OP2-OC2= PC2, but 0A2= OP2; AC2 = PC2 or AC = PC; hence the section APBA is a circle. If the cutting plane pass through the centre, the radius of the section is evidently equal to the radius of the sphere, and such a section is called a great circle of the sphere. 3. The poles of any circle are the two extremities of that diameter or axis of the sphere which is perpendicular to the plane of that circle ; and therefore either pole of any circle is equidistant from every part of its circumference, and, if it be a great circle, its pole is 90° from the circumference. A spherical triangle is the portion of space comprised between three arcs of intersecting great circles. 4. The angles of a spherical triangle are those on the surface of the sphere contained by the arcs of the great circles which form the sides, and are the same as the inclinations of the planes of those great circles to one another. 5. Any two sides of a spherical triangle are greater than the third side. B 9 PPHERIC8. Since by Euclid XI. 20, any two of the plane angles, which form the solid angle at (), are together greater than the third, hence any two of the arcs which measure those angles must be greater than the third. 6. Since the solid angle at 0 (see fig. p. 3) is contained by three plane angles, and by Euclid XL 21, these are together less than four right angles, hence the three arcs of the sphe¬ rical triangle which measure those angles must be together less than the circumference of a great circle, that is a-{-S + c> 360, and since any two sides of a triangle is greater than the third, we have 6 + c>a; a b.

ON THE POLAR OR SUPPLEMENTAL TRIANGLE. 7. If three arcs of great circles be described from the angular points A, B, C, of any spherical triangle A B 0, as poles, the sides and angles of the new triangle, D F E, so formed will be the supplements of the opposite angles and sides of the other, and vice versa. Since B is the pole of D F, then B D is a quadrant, and since C is the pole of D E,f C I) is a quadrant; therefore the distances of the points B and C from D being each a quadrant, they are equal to each other, hence D is the pole of B C. n DE = 180°—C; EF = 180o-A; AFD = 180°—B; and D = 180o-BC; E == 180°—AC; F == 180°—AB. , Also, AB=180o—F; BC = 180o—D; A AC = 180°—E; A = 180°- FE; . r-^E B = 180°—FD; C = 180°—DE. The sum of the three angles of a spherical triangle is greater than two right angles, and less than six right angles. For if a' -f b' + c' be the sides of the supplemental or polar triangle, A = ISO0—a'; B = 180°—6'; C == 180°—'c'; hence A + B-fC + a'+t'+c' = 6 x 90 = G right angles; but a'+ &'+ c is less than four right angles, by Euclid XL;, 21; therefore A + B + C is greater than two right angles; and as the sides a', b', c', of the polar triangle must have some magnitude, the sum of the three angles A, B, C must be less than six right angles. SPHERICAL TRIGONOMETRY.

CHAPTER I. 8. Spherical Trigonometry treats of the various relations between the sines, tangents, &c., of the known parts of a sphe¬ rical triangle, and those that are unknown; or, which is the same thing, it gives the relations between the parts of a solid angle formed by the inclination of three planes which meet in a point, for the solid ?. angle is composed of six parts, the incli- ; c nations of the three plane faces to each \ \ jj/1 "v"-- other, and also the inclinations of the ' X/ three edges; in fact, a work might be writ- \ / ten on this subject without using the / spherical triangle at all, for the six parts \ \ / of the spherical triangle are measures of the six parts of the solid angle at 0. See fig. 9. If a spherical angle have one of its angles a right angle, it is called, a^ right-angled triangle; if one of its sides be a quadrant, it is called a quadrantal triangle ; if two of the sides be equal, it is called an isosceles triangle, &c., as in Plane Trigonometry. 10. To determine the sines and cosines of a spherical tri¬ angle in terms of the sines and cosines of the sides. Let 0 be the centre of the sphere on which the triangle A B 0 is situated, draw the radii OA, OB, 0 0; from 0 A draw the perpendiculars A D and A E, the one in the plane 0 AB, and the other in the plane 0 AC, and suppose them to b 2 3: SPHEIUCAL TRIGONOMETRY. meet the radii OB and 0 0 produced in D and E. The angle D A E is equal to the angle A of the spherical triangle, and taking the radius unity we have A D = tan C, 0 D = sec C, A E = tan b, 0 E = sec b. Then in triangles DAE and D 0 E we have O D2 + 0 E2 - 2 0 D . 0 E cos E 0 D = D EJ AD2 + AE2 —2AD . AE cos A = DE2 by subtracting the second equation from the first, observing that OD2— AD2 = OE2— AE2= 1, and EOD is measured by B C or a, we obtain 2 + 2 A D . A E cos A — 2 0 D . 0 E cos a — 0; or by substituting the above values 1 + tan b . tan c cos A — sec b sec c cos a = 0 , 1 , sin b but sec b7 - j-, tan o = : cos 6 cos b 1 sin c sec c , tan c = : cos c cos c sin b sin c cos A cos a 1 H cos7 b cos c cos bj cos= c 0; hence cos a = cos b cos c + sin b sin c cos A (1) which is the fundamental formula in Spherical Trigonometry. 11. In the figure the sides b and c are less than 90°, but it is easily seen that equation (1) is gene¬ ral. Let us suppose that one of the sides, A C or b for example, is greater than 90°; draw the semi-circum- j.b ferences C A 0', 0 B C, and make the v ... - triangle ABC of which the sidea c7 a' and b', or B C and A C, are sup- •*'" plements of a and b, and the angle: B A C the supplement of A. Since the sides b and c are less than 90°, the equation (1) can be applied to the triangle A B C, and gives cos a' = cos b' cos c -f- sin sin c cos B A C ... (2). SrHERICAIi TRIGONOMETRY. Now a! = 180°- o, V = 180°-b, B A C = 180°- A; these values substituted in eq. (2) will give eq. (1), which shows that it is true for the case where b is greater than 90°. Let us now suppose that the two sides b and c are both greater than 90°; produce AB and A C till they b, / ; c intersect in A', which forms the c ./ / ' triangle B C A' in which the angle A' / is equal to A, and the sides b' and c' / B the supplements of b and c; by making the substitutions in this case, we still find that equation (1) satisfied. Lastly, we can verify equation (1) in the case where I = 90° and c = 90° either both together or separately. If we apply equation (1) to each of the sides of the triangle, we shall have three equations by means of which we can always find any three parts whatever of the triangle, when the three others are given. But, for practice, it is necessary to have separately the divers relations which exist between four parts of the triangle taken in every possible manner. There are in all four distinct combinations, which we proceed to give. 12. 1st, Relation between the three sides and an angle. By applying equation (1) to the three angles, we have cos a = cos b cos c + sin 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 -f sin a sin b cos C (3) 13. 2nd, Relation between two sides and their opposite angles. From equation (1) we have . cos a — cos b cos c cos A = ;— —;— sin o7 sin c s tt . 2. A - 22A . (cos a — cos b cos c) Hence sin A = 1 — cos A = 1 — . «4 , . » ~ sin' o sm-'c (1 — cos2 6) (1 — cos2 c) — (cos a — cos b cos c)3 sin2 b sin2 c 6 . SPHERICAL TRIGONOMETRY. sin A yi — cos2 a — cos2 b — cos2 c + 2 cos a cos b cos c sin a sin a sin 6 sin c We must take the radical with the positive sign, seeing that the angles and the sides are less than 180°; their sines are positive. As the second member remains constant when we change A and a into B and b, &c., we have sin A sin B sin 0 sin a sin b sin c Hence in any spherical triangle, the sines of the angles are to each other as the sines of their opposite sides. 14. 3rd, Relation between the two sides and their included angle, and the angle opposite one of them. In considering the combination a, A, 0; first eliminate cos c, between the equations (1) and (3) and we have cos a = cos a cos2 b + cos b sin a sin b cos 0 + sin b sin c cos A transposing cos a cos2 b, and observing that cos a—cos a cos2 b = cos a sin2 b; and, dividing the whole by sin b sin a, it becomes cos a sin b , ^ , sin c cos A ; = cos b cos 0 H ; , sm a sin a sin c sin 0 , . , , . but —r-; and consequently we have for the relation sin a sin A sought cot a sin b = cos b cos 0 + sin 0 cot A. By permuting the letters, we have in all the following six equations: cot a sin b = cos b cos 0 + sin 0 cot A (5) cot b sin a = cos a cos 0 + sin 0 cot B (6) cot a sin c = cos c cos B + sin B cot A (7) cot c sin a = cos a cos B + sin B cot 0 (8) cot b sin c = cos c cos A + sin A cot B (9) cot c sin b = cos b cos A 4- sin A cot 0 (10) SPHERICAL TRIGONOMETRY. 7 15. 4:th, Relation between one of the sides and the three angles. Eliminate b and c from the equations (1) (2) (3) : to do this we have by the last article cos a sin & _ ^ sin c cos A ; = cos o cos (J — . , sin a sin a sin b sin B _ sin c sin 0 and since ——=-—- and ——=-— sin a smA sin a sin A we have cos a sin B = cos b sin A cos 0 + cos A sin C; and changing a and A into b and B, and vice versa, we obtain cos b sin A = cos a sin B cos 0 + cos B sin C. We have only now to eliminate cos b by the two preceding equations. We find after reduction the relation sought between ABO and a, which, applied to the three angles successively, will give the three equations cos A = — cos B cos 0 + sin B sin 0 cos a (11) cos B = — cos A cos 0 + sin A sin 0 cos b (12) cos 0 = — cos A cos B + sin A sin B cos c (13) 16. The analogy of these equations with the fundamental formula is striking, and conducts us to a remarkable conse¬ quence. Let us imagine a spherical triangle A' B' 0', of which the sides a' V c' are the supplements of the angles A, B, C; then from equation (1) we shall have cos a' — cos V cos c' + sin b' sin b1 cos A'. Now sin a'= am A, cos a = — A, sin U = sin B, &c., then — cos A = cos B cos 0 + sin B sin 0 cos A!. From this equation we find for cos A! a value equal but of a contrary sign to that which we find for cos a in equation (11); then a = 180° — A', similarly b — 180 — B', and c = 180 — C. Hence, having given any spherical triangle, if we describe another triangle,, the sides of which are the sup¬ plements of the angles of the first, then the sides of the first will be the supplements of the angles of the second. From this property the two triangles are called supplementary, and eometimes the triangles are said to be polar to each other. 8 SPHERICAL TRIGONOMETRY.

NAPIER S ANALOGIES. 17. We now proceed to deduce the formulse known by the name of the analogies of Napier, which are employed to sim¬ plify some of the cases of spherical triangles. The equations (1) and (2) give cos a — cos b cos c = sin b sin c cos A; cos b — cos a cos c = sin a sin c cos B. By division, observing that S^n a ^ ^ sin b sin B ' we have cos b — cos a cos c sin A cos B cos a — cos b cos c sin B cos A * By subtracting and adding unity to both sides of this equation and again dividing cos b — cos a 1 -f cos c sin (A — B) cos b + cos a 1 — cos c sin (A + B)" But by Plane Trigonometry, page 30, (12), cos b — cos a = tan (a 4- b) tan J (a — b) cos b + cos a but1 + 003 c 1 — cos c tan2 ^ c and sin (A + B) = 2 sin % (A + B) cos J- (A + B) sin (A — B) = 2 sin ± (A — B) cos J (A — B) Substituting these values, the above equation becomes tan J (a + b) tan (a — b) = tan2 i ccos j(A-B)\_ \8in J (A + B) cos | (A + B) / (a) , . sin a sin A ana since-:—-—- sin b sin B we have sin a + h _ sin A + sin B sin a — sin b sin A — sin B* By Plane Trigonometry, page 30, tan | (q + b) sin | (A + B) cos | (A — B) tan j (a — b) cos £ (A + B) sin | (A — B)' SPHERICAL TRIGONOMETRY. 9 Multiply th^se two equations together, and then dividing one by the other and extracting the root, observing that tan £ (a + 5) and cos J (A + B) ought to have the same sign,

t™ + ^ ~ g - (H)

<15) To apply these to the polar triangle we must replace a, b, c, A,B,0, by 180°-A, 180o-B, 180o-C, 180o-.a, 180o-b, and there results, + = (1C)

= (17) My able and talented friend, Mr. Eeynolds, of Chelsea Hos¬ pital, has sent me the following very neat method of deducing Napier's Analogies, which he says was communicated to him by Mr. Adams, the celebrated astronomer of Cambridge. , sin A sin B sin A + sin B LetT m = = = — ;——. sin a sm b sm a + sin b Then by the formula (11) page 7, cos A + cos B cos C = sin B sin C cos a = m sin C sin b cos a...(I) cos B + cos A cos 0 = sin A sin 0 cos b = m sin C sin a cos b...(2) Add (1) and (2), then {cos A + cos B} (1 + cos 0) = m sin 0 sin (o + b). Also sin A + sin B = m (sin a + sin b).. Dividing and reducing we have a — b 10 SPHERICAL TRIGONOMETRY. Again, subtracting (2) from (1) (coa B — cos A) (I — cos 0) = vi . sin (a — b) sin 0 and sin B + sin A = m (sin a + sin 6); dividing a — b j^q sm 2 tan . tan _ = —— (Anal. 2) 8m —2— The other two follow of course from the polar triangle. ON RIGHT-ANGLED SPHERICAL TRIANGLES. 18. The preceding formulae will apply to right-angled triangles, if we make any one of the angles = 90°. If A = 90° we have cos a = cos b cos c (1) sin b = sin a sin B (2) sin c = sin a sin 0 (7) tanb = tan a cos C (3) tanc = tana cos B..;... (8) tan b = sin c tan B (4) tan c = sin b tan 0 (9) cos B = sin 0 cos b (5) cos C^= sin B cos c (10) cos a = cot B cot 0 (6) These six independent formulae are all adapted to logarith¬ mic calculation. The first gives a relation between the hypothenuse and the two sides containing the right angle; the second, one side and angle opposite; the third, between the hypothenuse, a side, and the adjacent angle; the fourth, between the two sides and the angle opposite to one of them; the fifth, between one side and the two oblique angles; lastly, the sixth, between the hypothenuse and the oblique angles. 19. The formula (1) requires that cos a must have the same sign as the product cos b cos c, or that the three cosines must be positive, or that only one must be so. Therefore in any right-angled spherical triangle the tnree sides must be less than 90°; or two of them must be greater than 90°, and the third less. The formula (4) shows that tan b has the same sign as tan B, and tan c the same sign as tan 0. Therefore each side containing the right angle is of the same kind or affection as the angle opposite, that is, the angle and the side are both less than 90° or both greater. SPHERICAL ^TRIGONOMETRY. 11

napier's circular parts. 20. As we have before observed, the above formulas are simple and well adapted for logarithmic computation, yet they are not easily remembered; therefore it is of importance that we should have some method which will relieve the memory as much as possible; this is supplied by what is termed'Napier's Circular Parts. By committing to memory the two rules which will be given hereafter, the student will he able to solve all the cases in right-angled triangles, as well as if he had all the formulas by heart. The circular parts of a right-angled spherical triangle are five, namely, the two sides, the complement of the hypothe- nuse, and the complements of the two angles (the right angle being always omitted). Three of these circular parts, besides the right angle, enter every proportion, two of which are given, and the third sought. These three parts are named from their positions with respect to one another, that is, according as they are joined or disjoined, observing that the right angle does not separate the sides. If the three circular parts join, that which is in the middle' is called the middle part, and the other two are called ex- tremes corijunct. If the three circular parts do not join, two out of the five must, and that part which is separate or alone is the middle part, and the other two are called extremes, disjunct.* These things being understood, the following is the general rule. The sine of the middle part is equal to the product of the tangents of the extremes conjunct. * Thus, if in figure page 12, we suppose B C, the angle B, and the fide A B to be the quantities that are to be used; now as they lie altoge¬ ther, the angle B is the midale part, and the two sides, B C and A B, are the extremes conjunct. Also, if the angle B, A B and A C be the quantities, then since the right angle does not separate the sides, A B is the middle part and the other two elements are the extremes conjunct. But if the quan¬ tities be A C, B C and the angle B, then the angle C is said to separate A C from B C, and the'side A B is said to separate A G from the angle B, that part A C which is separated from both the others, call the middle part, and the parts which are disjoined from it call extremes disjunct. This practical method will be useful to seamen, and requires very little effort of memory. 12 SPHERICAL TRIGONOMETRY. The sine of the middle part is equal to the product of the cosines of the extremes disjunct. From these two equations, proportions may be formed, observing always to take the complements of the angles and hypothenuse; and that the cosine of a complement is a sine, and the tangent of a complement is a co-tangent, and vice versa* c, 21. Case 1. When the hypothenuse B C and the base A B are given to find the /a* remaining parts of the triangle. Let us proceed to find A 0. c Here the hypothenuse and the two sides are the three circular parts. The hypothenuse being separated or disjoined from the sides it is the middle part, and the sides are the extremes disjunct. Then sin B 0 = cos A B cos A G. And since we must always take the complements of the hypothenuse and angles, this becomes cos a = cos h cos c. Now, as this agrees with equation (1), the rule is proved in this case. To find the angle B. Here the three circular parts all lie together, taking B to be the middle part, then A B and B 0 are adjacent parts, or extremes conjunct. .v sin B = tan B 0 . tan A B; taking the complements of B and B 0, we have, cos B = cot a tan c, which corresponds with (8), and therefore the rule is proved in this case also. To find the angle 0. Here the side A B is separated from the hypothenuse by the angle B, and it is separated from the angle 0 by the side A 0, then A B being the middle part, the hypothenuse and the required angle are the extreme or disjoined parts. sin A B = cos B 0 cos 0; taking the complements of B 0 and 0, sin c =.sin o sin 0. This agrees with (7), and therefore proves the rule. SPHERICAL TRIGONOMETRY 13 22. Case 2. Given the two sides &, and c, which include the right angle, to find the hypothenuse and the angles. 1. To find the hypothenuse. As the two sides are separated from the hypothenuse they will be extremes disjoined, the hypothenuse being the middle part; sin B 0 = cos A B cos A C; taking the complement of B 0, cos a = cos b cos c; which is the same as equation (1); cos c == cos =. a cos 6 To find angle 0. Since the right angle A does not disjoin, the three parts all lie together, hence A 0 being the middle part, A B and angle 0 are the adjacent parts, or extremes conjunct. sin A 0 = tan A B . tan 0 ; taking the complement of C, sin b tan c cot c; which agrees with (9); or tan, /nU = ——tan c Bin 6 To find the angle B. The three circular parts all lie together again, A B being in the middle ; calling it the middle part, then A C and angle B will be extremes conjunct. sin A B = tan A 0 tan B; taking the complement of B, sin c = tan b cot B, which agrees with (4); or tan, -oB = tan b sm c 23. Case 3. Given the hypothenuse a and angle B to find c, G. 1. To find AO or b. As A 0 is separated from the hypothenuse by the angle 0, and from the angle B by the side A B; calling A 0 the middle part, then B 0 and angle B are extremes disjunct. sin A 0 = cos B 0 cos B; taking the complements of BO and 0, sin b =sin a sin B, which is the same as equation (2). SPHERICAL TRIGONOMETRY* 2. To find A B or c. Here the three circular parts all lie together, and AB in the' middle ; calling it the middle part, then A B- and B C will be adjacent parts^ or extremes conjunct. sin B = tan AB tan B 0 ; taking the complements of B and B 0. cos B = tan c cot a; tanc= C0S = cos B tan a. which agrees with equ. (8). cot a -aw 3. To find G. Here the circular parts lie all together, the hypothenuse being in the middle ; call it the middle part, and the angles B and 0 will be adjacent parts. sin B 0 = tan B tan G; taking the complements throughout, cos a = cot B cot C ;■ cot C = = cos a tan B, which agrees with equ. (6). 24:. Case 4^ Given the side A C or b and the opposite anglQ B to find a, c, 0. 1. To find the hypothenuse B 0 or a. Here ^ or AO is separated from the hypothenuse by the angle 0, and from the angle B by the side AB; calling then A 0 the middle part, the angle B and the hypothenuse are the extremes disjunct. sin A 0 = cos B C cos B; taking the complements of B 0 and B, sin b ==sin a sin B which agrees with equation (2). sin a = sin. & . smB 2. To find c. Aa the right angle does not disjoin, c lies in the middle between b and B; calling it the middle part, b and B are the extremes conjunct. sin 0 = tan b tan B; taking the complement of B, eih c = tan b cot B, which agrees with equation (4). SPHERICAL TRIGONOMETRY» 15 3. To find C. Here the angle B is separated from c by the hypothenuse, and it is separated from b by the side A B; calling it the middle part, sin B = cos b cos C ; taking the complements of B and 0, cos B = cos b sin 0, which agrees with equation (5). cos B sin 0 : cos b There is here an ambiguity, since each quantity is deter¬ mined by its sine, and we see that this really ought to be the case. In fact, if the triangle BAG (fig. p. 12) right-angled at A, satisfy the equation; produce B A and B 0 till they inter¬ sect in D, then take D A' = B A, and D C7 = B 0, the triangles B A 0, D A' O7 will be equal in all respects, then the angle A is a right angle, and G' A! = OA = b. Thus the triangle B A'C is right-angled, and contains also the given parts B and b; we can therefore take at will a < 90°, or A > 90°, but when the choice is once made the affection or species of c will be determined by the equation cos a = cos b cos c, and that affec¬ tion will be the same as that of C. There will be only one triangle which has two right angles when b = B, and none when we have sin b > sin B. 25. Case 5. Given the side b and the adjacent angle C, to find a, c, B. 1. To find the hypothenuse a. Here the parts all lie together, the angle 0 being in the middle; call it the middle part, then A G and B 0 are the extremes conjunct. sin 0 = tan A 0 tan B C; taking the complements of 0 and B 0, cos C = tan b cot a; which agrees with equation (3); . . tan. a = tan—. b cos 0 2. To find AB or c. As the right angle does not disconnect, A 0 is the middle part, and A B and angle 0 are the extremes conjunct, sin A C = tan A B tan 0; 16 SPHERICAL TRIGONOMETRY. taking the complement of 0, sin b = tan c cot c; tanc =^= S^Q ^ == sin & tan C, which agrees with equ. (9) cotO r 3. To find the angle B. Since the angle B is separated from b by the side A B, and from the angle 0 by the hypothenuse BO; calling it the middle part, then A 0 and the angle 0 are the extremes disjunct. sin B = cos AO cos 0; taking the complements of B and 0, cos B = cos b sin 0, which agrees with equation (5). Here a, c and B are found without any ambiguity. 26. Oase 6. Given the two oblique angles B and 0 to find a, b,c. 1. To find a or B 0. Here a, B and 0 all lie together, a or B 0 being in the middle; call it the middle part, then B and C are the ex-' tremes conjunct^ sin B 0 = tan B tan 0; taking the complements of the whole, cos a = cot B cot 0, which agrees with equation (6). 2. To find & or A 0. Here Bis separated from0by the hypothenuse BO, and it is separated from A 0 by the side A B; calling it the middle part, then the angle 0 and A 0 are the extremes disjunct. sin B == cos 0 cos A 0 ; taking the complements of B and 0, cos B = sin 0 cos &, which agrees with equation (5). 3. To find c or A B. Here the angle 0 is separated from A B by the side A 0, and from the angle B by the hypothenuse B 0; calling it the middle part, then A B and the angle B are the extremes disjunct. sin 0 = cos A B cos B; taking the complements of 0 and B, cos 0 = cos c sin B, which agrees with equation (10); SPHERICAL TKIGONOMKTRY. 17 cos 0

These values leave no ambiguity, and if the triangle is impossible they will show that it is so. 27. When a'triangle is isosceles, the two equal sides are only counted as one element, and the angles which are oppo¬ site to them also as only one element. Now, if we draw the arc of a great circle through the vertex of the triangle and the middle of the base, we divide it into two right-angled triangles, equal in all respects, and in each of which we know two elements besides the right angle, then the isosceles triangle can be solved by the i'ormulai for right-angled triangles. 28. If in a spherical triangle ABC, in which we have a + 6 = 180°, produce a and c till they intersect in D, we shall have a + CD = 180°, hence CD = 5; therefore, the solution of the triangle A B C is brought c to that of the isosceles triangle A B D. The same thing may be said of a triangle, in which two angles are the a <-• spherical supplements of each other, for we cannot have A -f- B=180o without at the same time having A + B —180° and vice versd. In fact, in the isosceles triangle A CD, the angle CAD=D = B. Now, CAD + CAB = ISO0; then also, in the triangle A B C we ought to have A + B = 180o. 18 SPHEKICAL TKIGONOMETRV.

CHAPTER II. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES.

29. Case 1. Given the three sides a> b, c to find the angles A, B, 0. To find A, we have by equation (1) . cos a — cos b cos c COS A = ; = ; > sin o sin c but we obtain an expression better adapted to logarithms by finding sin J A, cos % A, &c., as in Plane Trigonometry. Since 2 sin2 i A = 1 — cos A, we have by substituting the above value of cos A, _ . - . ^ cos a — cos b cos c 2 sm20 i A 1 :— —; ^ sin u7 sin c cos b cos c + sin b sin c — cos a sin b sin c cos (b — c) —cos a sin b sin c (by equation (8) page 30, Plane Trigonometry,) 2 sin ^ (a + 6- c) sin -J- (a — b + c) sin b sin c 5

sin I A = . /Bi'1 i (« + b — c) sin j (a — 6 + c) L V sin b sin c For the sake of abridgment, put a + 6 + c=:2s, and the preceding expression becomes

/a ( ) n( c) »A=vV "sin'T\ o °sin c'- . In the sameme way

iA== /Bi'iSBin (s-q). cos V sin b sin c _ . /sin (s— b) sin (5 — cV . tan i A = A / — > —v v ! V sin 5 sm (s — a) SPHERICAL TRIGONOMETRY. * 19 30. Case 2. Given the two sides a, 5, and the angle A op¬ posite to one of them, to find c, B, 0. We obtain at first the angle B opposite to b by the proportion sin a : sin b : : sin A : sin B ; . sm. o7 =sin A sin .b sin a It will be best to determine c and 0 by Napier's Analogies, which give tan i c = tan i (a — b) Bin^(A + E) ^ 2 V ; ein£(A — B) cot i 0 = tan i (A — B) . 6inf (« + &), sin i (a — b) As the angle B is determined by its sine, it can either be acute or obtuse. However, for certain values of the given quantities a, b, A, there will be only one triangle. We may refer back to the similar case of plane triangles; we can thus find 0 in a direct manner by the equation cot A sin 0 + cos b cos 0 = cot a sin b. To effect this, let us at first determine an auxiliary angle 0 by putting cot A = cos b cot 0, from whence we have eot6=e^A; cos b then in the equation (5), p. 6, cot A=cos b cot 0= cos? CQS ^ sin $ the equation becomes cos b (sin 0 cos

The side c can also be obtained by equation (7), cot a sin c — cos B cos c = cot A sin B, in which we make cot a = cos B cot 0 22 SPHERICAL TRIGONOMETRY. . „ . . cot a . , ,v tan B sin 0 . . cot0 = sin (c.— 0) = — t * cosB tan A Lastly, we can find C, for sin a : sin c :: sin A : sin 0, or better by means of the equation, cos a sin B sin 0 — cos B cos G = cos A, we reduce the first member to a monomial by putting cos a sin B = cos B cot 0, from whence we have cot, 0=i cos a tancos B, sin (0 — 0) = A v»sin— 0 > v ' cos B these values determine 0, 0—0,and consequently the angle 0. 34:. Case 6. Given the three angles A, B, 0, to find the sides a, b, c. This case is solved in a similar way to the first. By page 7, equation (11)

cos a cos a + cos B cos C = sin -pB sm. nC > and by the same method, as used in the first case,

sin | a^ /si'iSsin^A-S) V sin B sin U • cos i a = /sin (B - 8). ein (G--S) AV sin B sm 0 tan J a = a // -^-7^sin S sin (A\ — S)7 >.-' V sin (B — S)gx sm (0/n — Qo) By using the polar triangle in Case 1, we have . 1 / — cos S . cos (S — A) sm a = a / . t> • n V sin B sm 0 /cos (S-B) cos (S —0) COS -h a = x / ^ ; =r^ ^ J V sm B sm 0 - /— cos S . cos (S — A) tan & a ios (S _ B) cos (SC)

The first and last of these appear under an impossible form, but since S is always greater than 90 and less than 270, the cos S is alwavs negative, and therciore makes the quan¬ tity under the radical always positive. SPHERICAL TRIdONOMETRY. 23

ON THE AMBIGUOUS CASES OF SPHERICAL TRIANGLES. 35. The only cases in which there is any uncertainty are the second and fifth. We proceed to show in this article what conditions are necessary that there may be two solutions, or only one, or even when the triangle is impossible. Let us consider upon a sphere a semicircle D 0 D' perpen¬ dicular to a whole circle 1) H D'; take G D less than 90° and draw the arcs of great circles OB, CB', C H. . . . from the point 0 to the different points of the circumference D H D'. Produce 0 D, making 0' D = 0 D, and join 0' B. The triangles C D B, CD B have a right angle contained between the equal sides, therefore C B = C B. Now we have 0 D C < 0 B + B 0', therefore 0 D < C B. Hence, in the first place, the arc C D is the least that we can draw from the point C to the circumference D H D'; and consequently 0 D' is the greatest. LetDB'=DB; then in the two triangles CD B and CDB' have the two sides C D, C B and the right angle C D B of the one, equal to the two sides C D, D B, and the right angle C D B' of the other, hence C B' = 0 B. Therefore, in the second place, the oblique arcs equally distant from C D or C D' are equal. Lastly, let D H > D B; draw C H and produce C B till it intersects C H in 1. Then, since the arc 0 C is less than a semicircle, it will meet 0 B produced beyond the point C; this requires that the intersection I falls between H and O. We have therefore C B < CI + I B, and consequently O B + B C < C I + 10. But we have I 0 < IH + H C, and therefore 0'I + IC<0'H+H0; hence, fortiori, C B + B 0 < C H + H C. Now, C'B ==B 0 and C'H = H 0, therefore we.have B 0

^ C is U the angle A by Art. 19; then, when A is acute, C D is the ghortest distance from the point 0 to the semi-circumference A E, and it is the greatest when A is obtuse. In the first hypothesis the triangle will be impossible if we have a < C D, which gives sin a < sin C D, and in the second it will be impossible if we have a > C D, which gives again sin A > sin 0 D. Now, in the right-angled sj)herical triangle A CD, we have sin 0 D = sin b sin A; then, in both hypotheses we shall have sin a < sin b sin A. On the other hand, when we seek the angle B of the unknown triangle ACB, we have sin. B-D =sin b; sin A; sin a then this value of sin B will be > 1, which is impossible. If we have a = 0 D, there will be only one right-angled triangle, A C D, which will be possible, and it is that which again indicates the value of sin B, which becomes sin B=l. It is understood that the angle A is not equal to 90°. Let us now examine the different relations of magnitude which the given quantities a, b, A can represent. Let A < 90° and b < 90° (fig. p. 23). Since A and b are < 90°, A D is also < 90° by Art. 19 ; then A D D E ; if now we have besides a b, it is clear that we can place between C A and C D an arc 0 B = a, and that on the other side, between 0 D and OE, we can put another C B' = C B = a ; that is to say, there are two triangles ACB and ACB' which have the same quantities given, viz., a b A. When a — b, the triangle ACB disappears, and there re¬ mains only the triangle A C B'. When a -f- b = 180, or when a + b > 180, the point B' coincides with E, or passes beyond it, and then no triangle can exist. SPHERICAL TRIGONOMETRY. We can discuss in the same manner the other ^hypotheses. The results are all contained in the following table. The sign ^ signifies equal to or greater than; and the sign eignifies equal to or less than. a<& two solutions, b <90° { a>6 one solution, a + &>1800 no solution. a + 5<180o two solutions, A<90°- 0 b>90° J a + &>180 one solution, a%b no solution. o two solutions, ) = 90 | a>6 no solution. a + 6 > 180° two solutions, b <90° a + b < ^180° one solution, a < Cb no solution. o two solutions, A>90 - o a &>90 ^Cb one solution, a-\-b :i8oo no solution. ( ja>5 two solutions, & = 90 no solution. a^b one solution, b <90° a<& no solution, a + b ^ b no solution. A == 90° \ f «<6 one solution, b >90° no solution, I a + 6 <180o no solution. > = 90° | a = 90° solutions ad infinitum. a<[or >90° no solution. By the properties of thfc polar triangle, we can apply the results to the fifth case, where A, B, a, are given, only taking care to change a, b, A into A, B, a, the sign > into <, and the sign < into j>. When the given quantities fall in a case where we ought to have only one solution, the calculation will still indicate two. But to discern which ought to be taken, it is sufficient to observe; that the greater angle must be opposite to the greater side, and conversely. See Lefebure De Fourcy's Trigonometry. 2G SPHERICAL TRIGONOMETRY. Suppose,Tor example, that we have given A=112°, a—102°, b = 106°. In the preceding table, among the cases which correspond to A > 90°, we consider that where b 90°, and among these that where a 180°, we conclude from the table that there is only one solution, and since b is > a, the angle B is greater than A, therefore B is obtuse.

TO REDUCE AN ANGLE TO THE HORIZON.• 36. Let B A C be an angle in an inclined plane, and AD the vertical passing through A. Draw the horizontal plane meet¬ ing the lines A B, A 0, A D, in E, F, G; the angle E G F is the horizontal projection of the angle B A 0, or, in other words, it is the angle BAG reduced to the horizon. It is this angle . EFG that we have to calculate, supposing the angles BAG, B A D, 0 A D, to have been determined by an instrument. The geometrical construction is easy, for the line AG being arbitrary, we shall have sufficient quantities given to con¬ struct at first the right-angled triangle, E A G and FAG, then the triangle E A F, and, lastly, //'d EGF. The calculation of the angle E G F is equally easy. If we describe a / ft 1 sphere from the centre A with any radius, the / L lines AB, AO, AD, where they meet the \ /y' sphere, will determine a spherical triangle j BOD, of which the sides are known by means of the given angles, and of which the angle BDC of the triangle is equal to the required angle EGF. Then by the first case of oblique-angled spherical triangles, page 18, we have •

sintA= /ein(*-\)s.in-

37. The following properties of spherical triangles we shall premise before entering on the numerical solution of triangles. Any side of a spherical triangle is less than a semicircle, and any angle is less than two right angles. For the limit of any plane angle is two right angles, and this is also the limit of any plane face of a solid angle. The sum of the three angles is greater than two right angles and less than six right angles. If the three sides of a spherical triangle be equal, the three angles will also be equal, and vice versd. If the sum of any two sides of a spherical triangle be equal to 180°, the sum of their opposite angles will also be equal to 180°, and vice versd. If the three angles of a spherical triangle be all acute, all right, or all obtuse, the three sides will be accordingly all less than 90°, all equal to 90°, or all greater than 90°, and vice versd. The sum of any two sides is greater than the third side, and their difference is less than the third side. The sum of any two angles is greater than the supplement of the third angle. The sum of the three sides is less than the circumference of a great circle. If any two sides of a triangle be equal to each other, their opposite angles will be equal, and vice versd. Since . cos a — cos b cos c cos A =* :—;—: , sin b sin c cos b — cos a cos c COS B = : ; # sin a sm c o 2 28 SPHERICAL TRIGONOMETRY. . . cos a — cos a cos c It b — a, these expressions are each = ; : sin a sin c cos A = cos B or A = B; that is, the angles at the base of an isosceles triangle are equal, and the converse of this holds also. From this it is easily shown that the greater side of a sphe¬ rical triangle is opposite the greater angle, for let ABO be greater than CAB, and make the angle A B D equal to the angle DAB; DA = D B. AC = AD + DC = DC4-r)B,butDC + DB>BC A 0 > B 0.

OX THE NUMERICAL SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. 38. When the hypothenuse and one side are given. Ex. 1. Given the hypothenuse B C = 63° 56' 7", and the side A B = 40°, to find the remaining parts of the triangle. To find the other side, A 0. Here the hypothenuse and the two sides are the three circular parts. The hypothenuse being separated or disjoined from the sides by the angles is therefore the middle part, and the sides the extremes disjunct. sin B C = cos A B cos A 0; taking the complement of hypothenuse as directed by the rule, cos B 0 = cos A B cos A C; log cos B 0 = log cos A B + log cos A C — 10 log cos A 0 = log cos B 0 — log cos A B + 10 = log cos,G30 5G' 7"— log cos 40°+10 = 9-64284G4: - 9-8842540 + 10 = 9,7585924; AO = 54° 59' 59"'G. SPHERICAL TRIGONOMETRY. 29 The side A 0 is acute, because-the hypothenuse and the given side have the same affection. To find the angle B. This angle connects the hypothenuse and the given side, and is therefore the middle part, and the other the extremes conjunct. sin B = tan A B . tan BO; taking the complements of the angle and hypothenuse, cos B = tan A B . cot B 0; log cos B = log tan A B + log cot B C — 10 = log tan 40° + log cot 63° 56' 7" • = 9*9238135 + 9-6894:258 -10 = 9-6132393; /. B = 65° 46' 5". The angle B is acute, as the hypothenuse and given side are of the same affection. To find the angle 0. Here the side A B is separated from the hypothenuse by the angle B, and it is separated from the angle ,0 by the side A 0; take it to be the midcjle part, then B 0 and the angle A are extremes disjunct. sin A B = cos B 0 . cos 0; taking the complements of hypothenuse and angle 0, sin A B = sin B 0 sin 0; log sin A B = log sin B 0.4- log sin 0 — 10, log sin 0 = log sin A B — log sin B 0 +10 = log sin 40° — log sin 63° 56' 1" + 10 = 9*8080675 + 0*04:65794 by taking comp. log. 63° 56' 7"; = 9-8546469; 0 = 45° 41' 21". The angle 0 is acute, the hypothenuse and given side being - of the same affection. 30 SPHERICAL TRIGONOMETRY. \ When the two sides are given. Given the side AO = 52° 13', and the side A B =? 42° 17', to find the remaining parts. To find the angle B. (See fig. p. 12.) As the right angle does not disjoin, A B is the middle part, and the angle and side A 0 are extremes conjunct. sin A B = tan B . tan A 0; taking the complement of B, sin A B == cot B tan A C ; log cot B = log sin A B — log tan A 0 + 10 = 9-8278843 + 10 - 10-1105786 = 9*7173057, which is the log cot 62° 27'; /. B = 62° 27', which is acute, like its opposite side. To find the angle 0. Here A 0 is the middle part, and the angle 0 and AB are extremes conjjinct. sin A 0 = tan A B tan 0; • taking the complement of 0 sin A 0 = tan A B cot 0; log cot 0 = log sin A 0 — log tan AB + 10 = 9-8978103 - 9*958754:2 + 10 = 9*9390561, which is the log cot of 49°. The angle is acute like its opposite side. To find the hypothenuse B 0. The hypothenuse being separated from the sides by the angles, it is the middle part, and the sides are the extremes disjunct. sin B 0 = cos A B . cos A 0; taking the complement of the hypothenuse, cos B 0 = cos A B cos A 0; SPHERICAL TRIGONOMETRY. 31 log cos B 0 = log cos A B -f- log cos A 0 — 10 = 9*8691301 + 9-7872317 - 10 = 9-6563613, which is the cosine 62° 31', which is less than 90°, A 0 and B 0 being alike.

When a side and its opposite angle are given. Given the side A 0 = 55°, and its opposite angle B = 65° 46' 5", to find the remaining parts of the triangle. To find the other angle 0. Here B is the middle part, being separated from A C by A B, and from the angle 0 by B C; .*. A 0 and 0 are the extremes disjunct. sin B = cos A 0 cos C; taking the complements of B and C, cos B = cos A 0 sin C; log sin 0 == log cos B — log cos A 0 + 10 = 9-6132407 + comp. log. 0-2414087 + 10 = 9-8546494 = log sin 45° 41' 21" /. C = 45° 41' 21".

The angle C is ambiguous; as it cannot be determined by the data alone whether A B, C, and B 0 are greater or less than 90°. To find the side A B. Here AB is the middle part, AC and B the extremes conjunct. sin A B = tan A 0 tan B; taking the complement of B, sin A B = tan A 0 cot 3 ; log sin A B = log tan A 0 + log cot B — 10 = 10-1547732 + 2-6532976 -10 = 9*8080708, which is the sin 40°; A B = 40°. 32 SPHERICAL TRIGONOMETRY. The side AB is also ambiguous for the same reason as above. To find the hypothenuse B 0. The side A 0 is the middle part, and B 0 and B are the extremes disjunct. sin A 0 = cos B 0 . cos B; taking the complements of hypothenuse and angle, B, sin A 0 = sin B 0 sin B; log sin A C = log sin B 0 + log sin B — 10 log sin B C = log sin A 0 — log sin B + 10 = 9*913364:5 + 0*04:00568 + 10 = 9-9534:213; B 0 = 63° 56' 7".

When a side and its adjacent angle are given. Given the side A 0 = 54/ 4:6', and its adjacent angle 47° 56', to find the remaining parts. To find the side A B. Here the circular parts all lie together, hence A 0 is the middle part, and A B and 0 the extremes conjunct. sin A 0 = tan A B tan 0; taking the complement of 0. sin A 0 == tan A B cot 0 ; log A 0 = log tan A B + log cot 0 — 10 log tan AB = log sin A 0 — log cot 0 + 10 = 9-9121207 - 9-9554535 + 10 = 9*9566672 which is the tangent of 42° 8' 46"; " /. AB = 420 8'46/', which is acute, like its opposite angle. To find the angle B. Here B is separated from the two given quantities; calling it the middle part, then A 0 and 0 are the extremes disjunct. SPHERICAL TRIGONOMETRY. 33 sin B = cos A 0 cos 0; taking the complements of B and 0, cos B = cos A C sin 0; log cos B = log cos A 0 + log sin 0 — 10 = 9-7611063 + 9-8706179 - 10 = 9-631724:2, which is cos 64:° 38' 31"; •. B — 64:° 38' 31". To find the hypothenuse B 0. Here the circular parts all lie together, and 0 being in the middle, is the middle part, and B 0 and A 0 the extremes disjunct. sin 0 = tan B 0 tan AO; taking the complements of the hypothenuse and of angle 0, cos 0 = tan A 0 cot B 0; log cos 0 = log tan A 0 + log cot B 0 — 10; log cot B 0 = log cos 0 — log tan A 0 + 10 = 9-8260715 - 10-1510145 + 10 = 9*6750570, which is the cotangent o£ 64:° 4:0' 34/'; .-.BO = 64:° 40' 34//.

QUADKANTAL TRIANGLES. 39. Quadrantal triangles can be solved by the same rules as right-angled triangles for using the polar triangle; we see that since one side is a quadrant, and that in the polar triangle A/= 180 — a; .-. A'=180° - 90° = 90. In the polar -triangle, since A' have by the equa¬ tions, page 10, cos a' = cos b1 cos d sin b1 = sin a' sin B' sin c' = sin a1 sin C; tan b1 = tan a' cos C tan cf = tan a' cos B' o 3 34 SPHERICAL TRIGONOMETRY. tan V = sin c' tan B' tan c' = sin V tan 0' cos B' = sin 0' cos V cos 0' = sin B' cos d cos a' = cot B' cot c' From these by substituting these values a! = 180° - A; V = 180° - B; c' = 180° - 0; A' = 180° - a; B' = 180° ~ 6; 0' = 180° - c; we get these results, cos A = — cos B cos C sin B = sin A sin b sin 0 = sin A sin b tan B = — tan A cos c tan 0 = — tan A cos I tan B = tan b sin c tan 0 = sin B tan c cos b — sin c cos B cos c = sin b cos 0 cos A = cot b cot c Or without using the polar triangle, . cos a — cos b cos c cos A = —— make a = quadrant, sm: o sin c ^ ' then cos a = 0, and we have » cos b cos c cos A —-— cot b cot c: sin: 6 sm: c ' i} cos b — cos a cos c cos b cos B = sm a sm c sm a cos c — cos a cos b cos c cos 0 = sin a sin b sin b From these equations, and the equation S?n ^ = s^n a e sin B sin V W can deduce all the cases of quadrantal triangles. Given A B = c = 32° 57' 6" and A 0 = b = 66° 32' to find B and A, ' cos A = — cot b cot c log cos A = log cot b + log cot c — 10 = 10-1882850 + 9-6376106 - 10 = 9*8258956, which is the cosine of 47° 57' 16" but since cos A is negative, A must be greater than 90°. SPHERICAL TRIGONOMETRY. 35

OBLIQUE-ANGLED TRIANGLES. 40. Case 1. Given the three sides, viz. A B = 79° 17' M" \ BO = 110° (to find the rest. AC =r 58° j To find the angle A. By page 18, sin J A = /sin - 6 ) sil1 ("-"). AV sin b sin c ' hence we have the following rule : From half the sum of the three sides subtract each of the two sides which contain the required angle. Add the log sines of these two remainders, and the arith¬ metical complement logs of the sines of the sides which contain the angle. Half the sum of these four logarithms will give the log sine of half the required angle. Thus: 79° 17' 14" 110 58 2)247 17 14 123 38 37 = J sum of the three sides. 79 17 14 44 21 23 first remainder log sin = 9,8445513 123 38 37 58 65 38 37 second remainder log sin = 9,9595173 comp log sin 58° 0,0715795 comp log sin 79° 17' 14/' 0,0076359 2)19,8832840 log sin = 60° 57' 28" = 9,9416420 2 121 54 56 equals the required angle A. 36 SPHERICAL TRIGONOMETRY. By a similar operation the angles B and 0 may be found; but when one angle is known, the other two are easily deter¬ mined by Art. 13, page 6. Case 2. Given the angle A = 32° 20' 30", the side 5 = 72° 10' 20", and the side a = 78° 59' 10", to find B, 0 and c. Herett ,by page • 6, sm Br, = sin A; . sin 6 sm a log sin B = log sin A + log sin b — log sin a log sin A = 9*7283269 log sin b = 9*9786283 19*7069552 log sin a = 9-9919261 log sin B == 9*7150291 B = 31° 15' 15" By page 9, equation (16) cot J O = tan J (A + B) CQS % (a + 2 v n J cos | (a — b) log cot J 0 = log tan § (A + B) + log cos J (a + b) — log cos J (a — b) log tan J (A + B) = log tan 31° W 52" = 9-7217470 log cos J (a +5) •-= log cos .75° 34/ 45" = 9*3962727 19-1180197 log cos -J- (a — 6) = log cos 3° 24/ 25" = 9*9992318 log cot J 0 = 9*1187879 ... jo = 82° 30' 39"; or O = 165° 1' 18". We might find c from the equation sin C sin c = sm a . -— , sm Ar but we can find it directly from Napier's Analogies. SPHERICAL TRIGONOMETRY. 37 By page 9, equation (14:), we have

tan . c = tan i(a + b). ;

log tan 50 = log tan J (a + 6) + log cos i (A + B) — log cos |(A — B) log tan J (a + 5) = log tan 75° 34/ 45" = 10-5898236 log cos | (A + B) = log cos 3104:7' 52" = 9-9293745 20-5191981 log cos i (A — B) = log cos 1° 5' 15" = 9-9999218 log tan 50 = 10-5192763

i c = 73° 10' 10" ' c = 146° 20' 20".

Case 3. Given 0 = 30° 45' 28"; a = 84° 14' 29"; b = 44° 13' 45", the two sides and the included angle, to find A, B, c.

By Napier's Analogies, page 9, equations (16) and (17),

tan J (A + B) = cot} C . 003 f ~ ^ 2 K ' 3 cos i (a + 6)

and tan i (A — B) = cot ^ 0 . 8laf{a~^ 2 v ^ sin 5 (o + 6) | 0 = 18° 22'44" log cot = 10-4785395 i (o - 6) = 20° 0° 22" log cos = 9-9729690 20-4515085 £ (a + 6) = 64° 14' 7" log cos = 9-6381663

log tan J (A + B) = 10-8133422 J (A + B) = 81° 15' 44" • 41. 3& SPHERICAL TRIGONOMETRY. A — B determined, J 0 = 18° 22' U" log cot = 104785395 J (a—b) = 20 0 22 log sin = 9*534:1789 20-0127184: J (a + 6) = 64: 14: 7 log sin = 9-9545255 10-0581929 J (A - B) = 48° 49' S8". A and B determined. c determined, i (A+B)=810 15' 44:,,-41 log sin 36° 45' 28/,= 9-7770158 !(A-B)=48 49 38 logsin44 13 45 = 9-8435629 A = 130° 5' 22"-41 19-6205787 B = 32 26 6-41 log sin 32 26 6 =*= 9-7294422 log sin c = 9-8911365 c = 51° 6' 121', c may be found directly, without finding A and B, by the following method:— cos c—cos a . cos ft ^ Since cos 0 = : :—7 , page 5 sin a . sin o t .\ cos c = cos a. cos b + sin a. sin 5. cos 0; but cos 0 = 1 — ver. sin C, cos c = cosa. cos b + sin a . sin 5—sin a. sin b. ver, sin = cos (a—b) —sin a . sin b . ver. sin C;

... 1—cos c, or 2 sin2-|'=ver.sin(a-—5)+sin a. sin b. ver.sin

. , / sin a . sin 6 . ver. sin 0 \ = ver- 8,n \ 1 + ver. sin (a—b) } sin a . sin b . ver. sin 0 ver. sin (a—6) SPHERICAL TRIGONOMETRY* 39 which in logarithms is 2 log tan 0 = log sin a+log sin 5+log ver. sin 0—log ver. sin (a—b)...[a] then 2 sin2-^- =ver. sin (a—b) . sec2 0, and log 2.+21ogsin-^-=log ver. sin (a—6)+2 log sec 0—10...[6] c computed independently of A and B. Finding the auxiliary angle 0 by the foam [a], a = 84;° 14! 29" sin = 9-9978028 5 = 36 13 45 sin = 9-8435629 0= 36 45 28 ver. sin = 9-2984762 29-1398419 o — 6 = 40 0 44 ver sin = 9*3693878 .-. 2 log tan 0 = 19-7704541 and log tan 0 = 9-8852270 Case 4. Given c = 50° 6' 20"; A = 129° 58' 30"; B = 340 29' 30"; to find a, b, 0. By equations (14) and (15), page 9, tan ^ (o -f- 6) ^ tan ^ c . 008 i 5} T ' 2 cos£(A + B) tan £ (a - b) = tan £ c . S|n f (A ~ B) ^ ' 2 sin £ (A + B) £ (A + B) = 82° 14' \ i(A - B) = 47° 44' 30" ( | c = 25° 3' 10") log tan i (a + b) = log tan J c +log cos|(A — B) — logcos|(A + B) log tan i c = log tan 25° 3' 10" = 9-6697162 log cos J (A—B) = log cos 47° 44' 30" = 9*8276758 19-4973920 log cos ^ (A + B) = log cos 82° 14' = 9-1307812 log tan J (a -f b) = 10-3666108 40 SPHERICAL TRIGONOMETRY*

••• i (a + 6) = 66° 4:4:' 10 log tan J (a—6) = log tan J c + log sin J (A — B) — log sin £ (A + B) log tan i c = log tan 25° S' lO" = 9*6697162 log sin|(A—B) = log sin 47° M' 80' = 9*8693023 19-5390185 log sin £ (A + B) = log sin 82° 14' = 9*9959977 log tan £ (a—5) = 9*5430208 i (a-b) = 19° 14' 50'7 J (a + b) + J (a — b) = a £ (a + b) — £ (a — b) == b 66° 44' 10" 19° 14' 50" 85° 59' 0" 47° 29' 20"

a = 85° 59' and b = 47° 29' 20".

To find 0. sin A sin G sin a sin c . « . sin A or, sin U = sin c . ~— sin a log sin 0 = log sin c + log sin A — log sin a log sinA=log sin 129° 58' 30" = 9*8844129 log sin c = log sin 50° 6'20" = 9*8849241 19-7693370 log sin a = log sin 85° 59' = 9*9989319

Jog sin 0 = 9*7704051 /. 0 = 36° 6' 50". SPHERICAL TRIGONOMETRY. 41 Oase 5. Given the angles A = 70° 39'; B = 48° 36'; a = 89° 16' 53", to find the rest. By Art. 33, page 21, we have sin 6 = 8111 a log sin b == log sin a + log sin B — log sin A log sin a = log sin 89° 16' 53" = 9-9999658 log sin B = log sin 48° 36' = 9*8751256

19-8750914: log sin A = log sin 70° 39' = 9*974:7475

log sin 6 = 9*9003439 6 = 52° 39' r sin 5 =sin (180 — b) sin 127° 20' 56"; but since A > B, a must be greater than 6, hence b cannot be 127° 20' 56'/. To find c.

t By Napier's Analogies tan c = tan £(« + &). 003 f ^ ^ 2 v ; cos i (A — B) log tan J c = log tan i (a + b) + log cos | (A + B)—log cos i (A — B) log tan | (a + b) = log tan 70° 57' 59* = 10*4622011 log cos | (A + B) = log cos 59° 37' 30" = 9*7038563

201660574: log cos £ (A — B) = log cos 11° 1' 30" = 9-9919097 log tan | c = 10-1741477 £ c = 56° ir 29", or c = 112° 22' 58/,. By equation (16), page 9, we have

cot 10 = tani(A+B) ^ cos ^ (a — ft) ** SPHERICAL TRIGONOMETRY. log COt ^ 0 log = log tan J (A + B) + log cos J (a + b) — log cos J {a—b} log tan £ (A -f B) — log tan 59°- 37' 30" = 10-2320208 log cos i (a + b) log cos 70° 57' 59" = 9-5133811

19-74:54019 log cos £ (a — b) = log cos 18° 18' 54" = 9-9774233

log cot J 0 = 9-7679786-

^ 0 = 59° 37' 30", or 0 = 119° 15'.

Case 6. Given the three angles, angle A = 120° 54' 56" \ angle B = 50 >res^ angle 0 = 62 34 6 ) •

By page 22, cos * «= /cos (S-B) cosJS-C>; V sm B sm 0 hence the following rule.

To find the side B CL From half the sum of the three angles take each of the angles next the required side.

Add the log cosines of these two remainders, and the cpmp. log of the sines of each of the adjoining angles.

Half the sum of these four logarithms will give the cosine of half the required side; thus SPHERICAL TRIGONOMETRY. 43 121° 54:' 56" 50 62 34 6

2)234 29 2

117 14 31 62 34 6

54 40 25 first rem. cos 9,7621032 67 14 31 second rem. cos 9,5875321 comp. log sin 50° • 0,1157460 comp. log sin 62° 34' 6" 0,0518018 2)19,5171831 cosine 55° 9,7585915 2 110° = the required side B 0. By the same method the other sides may be found; but one side being known (with the angles) the rest are most readily found by Art. 14, page 6. 44 SPHERICAL TRIGONOMETBY^

CHAPTER III.

41. The surface of the sphere included between the arcs D M, D N is proportional to the angle N D M or the arc M N. See fig. page 1. If the circumference be divided into equal parts as M N, and great circles be drawn from D through the points M, N, the portions of the surface, such as N D M, are all similar and equal, hence if F M contains N M, n times, or if F M = n times N N, the surface F D M will be n times N D M. When D M coincides with D G, the angle FD G- or its mea¬ sure F M Gr =180; hence if S = whole surface of the sphere, and if A be the angle N D M which is measured by the are S A N M, the surface N D M = -j • but S = area of 4 great circles of the-sphere. Hall's Diff. Cal., page 370. S = 4 tt r2 = 4 tt when radius is unity, or y = tt = 180. 4 The measure of the surface of a spherical trangle is the difference between the sum of its three angles and two right angles. Let the triangle be A B C, a, 6," c, repre¬ senting the magnitudes of the angles at A,B,C;let P=surfaceBOiB,Q±=mCwm, R = AO n A; produce the arcs 0 m, 0 n, till they meet at e (which will be on the hemisphere opposite to that represented by e A B mnA)% then each of the angles at 0 and e equals the angle of the planes in which the arcs C m e, C n ey lie; therefore the angles at 0 and e are equal. SPHERICAL TRIGONOMETRY. 45

Again, the semicircles A 0 m, 0 m e; B C nt 0 n e are equa or, AG-f-OTTi = Cm + me,and/. A 0= me,and BO =nc aud the triangle men — the triangle A B 0; let x = its area then, by last article, x+ 2 180

*+Q=! • ^andz+P + Q + R^ a: + R=— • consequently, by addition, 2 * +

*' X = 4 (180) ^ ^ c)— j=a+&+c — 180°; or r* (a + 6 + c — 180°). Hence the area of a spherical triangle is equal to the ex¬ cess of the sum of its three angles above two right angles, which is called the spherical excess. The late Professor Woodhouse, in his able work on Tri¬ gonometry, observes that—" This expression for the value of the area was merely a speculative truth, and continued barren for more than 150 years, till 1787, when General Eoy employed it in correcting the spherical angles of observation made in the great Trigonometrical Survey." In a biographical sketch of the life of Isaac Dalby, late Professor of Mathematics at the Eoyal Military College, Sandhurst, in Leybourne's Mathematical Repository, it is stated that he had sent some years previously to his death an account of the principal events of his life after reaching maturity. The following is a quotation from himself given in the above-named excellent periodical:— "General Koy^ account of this measurement is in the Philosophical Transactions ; but it is not altogether what it ought to have been. His description of the apparatus, de¬ tail of occurrences, &c., are all well enough; but he should not have meddled with the mathematical part, for Ms know¬ ledge did not extend beyond Plane Trigonometry. I drew up 46 SPHERICAL TRIGONOMETRY. the computations in that form which I thought the most pro¬ per for publication, but he was continually making alterations. He didnot even understand the rule I made use of for finding the excess of the sum of the three angles of a spherical triangle above 180° (which since that timehas been quoted as General Roy's theorem), and would not insert it until he had consulted, the Hon. Henry Cavendish. For conducting the business in the field, however, few persons could have been better qua¬ lified than the General. . . " I believe he was the best topographer in England, and knew the situation of every barrow, cairn, and hillock in Great Britain. He had something of an observatory in the upper part of his dwelling, and could regulate a clock or watch by means of transits. In fact, he was ready enough at calculations which depended merely on the use of the tables. But the rules which he published for measuring the heights of the barometer all came from Mr. RamsdenV A note is given to this extract in the Repository, which is as follows:—" It is not until very recently that Mr. Dalby has had justice done him with regard to this ingenious rule. At page 138 of the new edition of Yol. III. of Hutton's Course of Mathematics, published by Dr. Gregory in 1827, we find this note :—' This is commonly called General Roy's rule, and given by him in the Philosophical Transactions for 1790, p. 171; it is, however, due to the late Mr. Isaac Dalby, who was then General Roy's assistant in the Trigonometrical Survey, and for several years the entire conductor of the mathematical department.'"

FURTHER DEVELOPMENTS CONCERNING THE SPHERICAL EXCESS. 42. Let the radius of the sphere be unity, tt the semi- circumference of a great circle; a, &, c, the three sides of a spherical triangle; A, B, 0, the arcs of a great circle that measure the opposite angles. Let the spherical excess A + B + C— 7r=S. The area of the spherical triangle is equal to the arc.S multiplied by the radius, and is therefore represented by S. Now, by Napier's Analogies, SPHERICAL TRIGONOMETRY. 47

tan I (A + B) = cot £ 0 . —^ - v 7 ^ cos f (a + 6)

f 1 / A T»\ ) ^an "f" "I" ^an "i" ^ tan (A + B) + J 0 } = 1 _-tan ^ (A+B) tan ^ c

cot i 0 . C0Sr I""?? + tan 4 0 cosJ(a + 6) 1—cot! 0 . C03|(a-°j . tan i 0 2 cos 5 (a+6)

cot i 0 . (a ~ H + tan 10 cos J (a + 6) ^ cosJ-(a — b) cos -g" (& -j- 5)

_cot4- 0 . cos J (a — 6) + tan ^ C . cos J (a + 6) * ~~ cos ^ (a + b) — cos J (a — 6)

1+cosC 1—cosO ,, —=—cos i{a—b) H . ^ cos i (a + o)N sm 0 2 v y sm 0 a \ » / cos % (a + 6) — cos -J- (a — 6) (by expanding the cosines and reducing)

1 cos J a cos | b + sin % a sin J b cos 0 sin 0 — sin ^ a sin J b

— cat ^ a cot | b — cos 0 sin 0 but tan f (A + B + 0) = tan i (S + 180) = — cot J S ;

^ „ cot i a cot i b + cos 0 cot J^ S = oin ( : 4:8 SPHERICAL TRiaONOMETRY. This equation, which is very simple, enables us to find the area of a spherical triangle when the two sides and the in¬ cluded angle are given. To find the area of a spherical triangle in terms of the three sides, , , ^ cot i a cot -J- b + cos 0 cot I S = = . ' sin 0 ^ cos c — cos a cos b COS 0 = ; ;— sin a sin 6

1-f cos a A ^ ^ l-^cos b and cot J o = T , cot J b oir*sin a" ' * 8111 b 1+cosa + cos6 + cosc COS 0 + cot -k a cot bb~ : : 7 * * sin a sin 6 _ . a4-6+c . a- h b c 2-sin sm - ^ cos c—cos (a+6) 1 + C0SC = : r—jr - sin a sin b sin a sin b 2sm(a + c—b) . b+c—a . _ cos (a—/ 6<7)—cos c ^-77 2 sin2 1 —COS C = ^^^ = ; : 7 sin a sin o sin a sm b

Multiplying these two equations and. extracting the root,

/To -4-6 + c . a-\-b—c . a-fc— b . 64-c—a A / sm —— sm sm — sm sin C = 2 sin a sin b

By substituting these values we have 1 + cos a + cos b + cos c cot -J- S = 2 4-&4-C . a+6—c . a+c—6 . 6+c a / . a 1 sin sm—-—sin- a / sm— 2 2 2 2

This solves the problem, but it can be put into a simpler form, and one that is adapted to logarithmic computation. SPHERICAL TRIGONOMETRY. 49 Kesuming the formula

f i q cot! a cot J 6 + cos 0 COt4b= siiTO ' l + cot2|S= 1 :sinT 2JS • cot2-Ja cot21-6 + 2 cot J a cot £ b cos O+l £in2 0 ' by multiplying botli sides of the value of cos 0 by 2 cot J a cot J 6, we have . 1 i i 7 COS C COS d COS 1} 2 cot I a cot ■£. b cos 0 = . , 2 sin2 -I a sin2 J 6 putting in the numerator for cos c, cos a, cos b, their values 1-2 sin2 J c, 1 — 2 sin2 Ja, 1 — 2 sin2 J 6, we shall have

2 cotiacot^ cos C = + _ 2 3 sm2^asin2|6 2

Also, cot2 -J-a cot2 J "■ 1~Sif^5 sin2fa sin22J6 1 — sin2 \ a — sin2 ^ b , ^ sin2 i a sin2 b b

Substituting all these values, we have

2 1 — 1 — sin \ c or sin2 J S sin? ^ a sin2 J b sin2 0'

. i r* sin -J- a sin i b sin 0 sin -I S = ^ cos c

Substituting for sin 0 its value, we have D 50 SPRTTRICAL TRIGONOMETRY. ein ^ S = . a + ^i-c . a + 6 —c. a + c--b . b + c—a>

2 cos a cos J 6 cos J c which expression is adapted to logarithmic computation. If we multiply this equation by cot J S we have

COS ^ S : 1 + cos a + cos b + cos c 4 cos ^ a cos J b cos J c cos2 ^ a + cos2 J b + cos2 J c—1 ^ 2 cos ^ a cos 6 cos J c ' From this we have, 1 — cos i S , . , q or Atan J S0 = sin J S 4 1—cos2*! a — cos2 J &—cos2-I c + 2 cos J a cos J 5 cos \ c

/ f . a+&+c . a+6—c . a+c—6 . 54-c—a") V t8m~2 'Bin 2 '8111 2 'sm 2 / The numerator of this expression can be put under the form (1—cos2 ^ a) (1 — cos2 J b) — (cos J a cos ^ b — cos J c)2 which may be decomposed into the factors sin J a sin ^ b + cos J a cos J b — cos J c and sin £ a sin \ b — cos % a cos J b + cos J c. These reduce ultimately, the first to cos (-J a —^ b)—cos J c=

20 sm. —a+c — 5s m. 5 + c— a 4 4 #' the second to cos J c — cos (J a + ^ &) = ^ . a-J-fr-j-c • & — c 2 sin , - sm —-t ; SPHERICAL TRIGONOMETRY. 51 tan i S = . . — C . Cl-J-C—— & » S-f-C"-® 4: sm am sin sm

/ ( . a+S+c . a + 6—c . a+c—b . 54-c—a) V rn 2 Sln 2 sm 2 8m / . tan £ S = a4-6+c a+b—c a+c—b &+c—a tan tan tan . tan • 4 4 4 4 This elegant formula is due to Simon Lhuillier. See Legendre's Geometry, page 319.

GIKARD'S THEOEEM. 43. By page 45, ISO0 a 4- b + c — 180° = r x, or reducing a, 6, c to seconds, TT T* 180° x 60 x 60 the excess m seconds = 5 x. TT H Now, on the earth's surface, the length of 1° taking a mean measurement = (60859*1) x 6 feet, and an arc = ' 360 radius = -— ; A TT 360 .*. (60859*1) x 6 x —— = radius of the earth in feet; A TT . , 2 ir 60 x 60 the excess m seconds = x . r-— • 360 62 x (60859-1)2 2s- x 10 2.x. 36 x (60859-1) n' log . excess = log x —12 (log 6 -4- log 60859-l)—log 2 tt x 10^ = log * - (11-1249536 - 1-7981799) = log x - 9-3267737. d 2 52 SPHERICAL TRIGONOMETRY. Hence the following rule:— From the logarithm of the area of the triangle, considered as a plane one, in feet, subtract the constant logarithm 9,3267737, and the remainder is the logarithm of the excess above 180°, in seconds nearly. Observed angles. Ex. Hanger Hill Tower . (а) 42° 2' 32" Hampton Poor-house, (б) 67 55 39 King's Arbour (c) 70 1 48 Distance from (a) to (b) = 3,84:61*12 feet, . from (a) to (c) = 24704:-7. Taking the distance from (a) to (c) for the base of the triangle, the perpendicular on the base will be 38461 *12 x sin 4:2° 2' 32//, and therefore the area of the triangle base perpendicular 2 = 24:704:-7 x 19230*56 x sin 42° 2' 32", log area = log 24704: 7 + log 19230*56 + log sin 42° 2' 32" - 10 = 4-3927761 + 4-2839906 + 9-8258661 = 8*5026328 = logarithm of the area in feet ; hence, 8*5026328 — 9*3267737 = — 1*1758591 ,* the corresponding natural number is *14992, the spherical excess in seconds. legendre's solution of spherical triangles whose sides ARE VERY SMALL COMPARED WITH THE RADIUS OF THE SPHERE. 44. When the sides a, b, c, are very small with respect to the radius of the sphere, the proposed triangle is very little dif¬ ferent from a rectilinear triangle, and, considering it as such, we can have a first solution approximately, but we neglect in this manner the excess of the sum of the three angles above 180°. To have a solution more approximate, we must take SPHERICAL TRIGONOMETRY. 53 into account this excess, and this we can do very easily by means of a general principle which we proceed to demon¬ strate. Let r be the radius of the sphere upon which the triangle is situated, and if we imagine a similar triangle upon the sphere whose radius is unity, the sides of this triangle will be -, —, —; and we shall have cos A = r r r a be cos — — cos - cos - r r r . b . c ' sm - sin — r r but since r is very great with respect to a, b, c, we shall have approximately, a a2

+ c2 _ a2 a4 ^^4 __ c4 &2 c2 2r2 24: r4 4:r4 cos A= —— b6c c /A 62 c2 \ r2 \ 6 r2 6 r2 / 2)2 i 2 Multiplyinglying numeratornumeral and denominator by 1 -j and 6 r2 reducing 54: SPHERICAL TRIGONOMETRY. COS A = b2 + c2 — a* a* + bA + c* — 2a2b2 — 2a2c2 — 2&2c2 2 6c 24: 6 c r2

Let A' be the angle opposite to the side a in the recti¬ linear triangle of which the sides are equal in length to the arcs a, 6, c, we shall have

cos A' — ^ +

from whence we have x = sin A', 6r2 6 c and since # is of the second order with respect to - and it r r follows that the result is exact to quantities of the fourth order, we shall then have *

A = A' + ^ sin A'; br2 but ^ b c sin A' = the area of the rectilinear triangle, of which the three sides are a, b, c, do not differ sensibly from those of the proposed spherical triangle. Then, if either area be called a, we shall have

A=A' + ^2) orA' = A-^;

Similarly, B'= B — ^; 0' = 0 — ^; hence there results, SPHERICAL TRIGONOMETRY. 55

A' + B' + C or 180 = A + B + 0 — -Jj, we "can then consider as the excess of the three angles of r the spherical triangle above two right angles. Hence we have the following rule :— A spherical triangle being proposed, of which the sides are very small with respect to the radius of the earth, if from each of its angles one-third of the excess of the sum of its three angles above two right angles be subtracted, the angles so diminished may be taken for the angles of a rectilinear triangle, the sides of which are equal in length to those of the proposed spherical triangle; or, in other terms :—The spherical triangle, whose sides are nearly rectilinear, of which the angles are A, B, 0, and the opposite sides a, b, c, answer always to a rectilinear triangle whose sides are of the sarm length, a, b, c, and of which the opposite angles are A —^ e , B — ^ t; 0 — -jc; e being the excess of the sum of the angles of the spherical triangle proposed above two right angles. The excess t, or wdiich is proportional to the area of the triangle, can always be calculated d 'priori, by the given parts of the spherical triangle considered as rectilinear. If the two sides, b, c, and the included angle A, are given, we shall have the area a = b c sin A; if we have given the side a, and the two adjacent angles B, 0, we shall have the area sin B sin 0 8in (B + C) * 45. Given the three edges of a parallelepiped, and the angles be¬ tween them, to find the solidity. Let the edges S A = /, 8 6 = #, SC = h, and the contained angles ASB = a, A SC = g, BSC = y, if from the point C we let fall C 0 perpendicularly on the plane ASB then in right-angled triangle OSO; CO = OS sin CSO = h sin 0 S 0, besides the surface of the 56 SPHERICAL TRIGONOMETRY. parallelogram A S B P =fg sin a. Therefore, if we call S the solidity of the parallelopiped S T, we shall have S = /. g. Ti. sin a sin 0 S 0. We now proceed to find sin OSO. From the point S as a centre and radius unity, describe a spherical surface meeting the right lines S A, SB, SO, SO, in the points D, E, F, G, we shall have a triangle D E F, in which the arc F G is perpendicular to E D, since the plane 0 S 0 is perpendicular to A S B. Now, in the triangle D EF, where the three sides, D E = a, D F = €, E F = y, we have ^ COS € — COS a COS y cos E = : : and sm a sm y sin E = — COS2 a — cos2 6 — cos2y + 2 cos a cos € cosy sin a sin y Then, in the right-angled triangle EFGr, sin G-F or sin OSO = sin E sin E F == sin y sin E, S = /. g. A. sin a sin y sin E = f.g.Ji. */1 — cos2 a — cos2 € — cos2 y + 2 cos a cos € cos y. The expression under the radical is composed of the two factors, sin « sin y + cos € — cos a cos y, and sin a sin y — cos € + cos a .cos y, the first cos € — cos (a + y) r= 2 sin a-f-€,4"Y . a -{- y — € 2 'Sin 2 ; the second == cos (« — y) — cos € = «'+€ — y . € + y — a 2 sin ^ • sin 2 ' therefore, S =

. a+£+y1 Sln. a+€—y Sin. «+y—€ 31. 11 €+y —« zfgh's/1sin g ~2— 2 —'

46. The same things being given as in the above to find the diagonal. SPHERICAL TRIGONOMETRY. 57 Let the diagonal of the base S P = z, and the required diagonal S T = u, the triangle A S P, iii which cos S A P = — cos «, we have z2 =f2 + g2 + 2fg cos a, in like manner in the triangle T S P, in which cos T P S = — cos 0 S P; u2 = z2 + h2 + 2 h z cos 0 S P. We must now find cos 0 SP, or of the arc F H. Now, in the spherical triangle EFH, we have cos F H = cos E F cos E H + sin E F sin E H cos E, substituting the • _ _ ^ COS € COS a cosy- , values F F = y and cos E = ; ; it becomes ' sm a sm y

€03 F^T HT = COS y COS E-»-• EMtt i sinEH; (cos. ^b — COS a COS y). = ' sm a sin E H cos € ^ sin (a — E H) cos y _ sin a sin a sin E H cos € + sin D H cos y sin a Therefore 2 hz cos F H, or 2 h z cos 0 S P =

20 h, cos -pb z sin E H+ , 20 ti, cos y .^ sin; D H: sin a sm a but in the triangle B S P we have

BF= 8 P.^P, .»dBS=S F »» »18, sin S B P sm S B P

which„ i • •i gives' n z sin E= H /r and, ,z sin D= H q: sm: a * sin a /. 2 h z cos 0 S P = 2 fh cos € + 2 g h cos y< Hence the square of the required diagonal • w2 ^ g2 ^ ^2 _J. 2 fg COS a + 2/A cos € + 2 g h cos y.

47. To determine a line on the surface of a sphere on which the vertices of all triangles of the same base and sur¬ face are situated. d 3 58 SPHERICAL TRIGONOMETRY. Let A B C.bea spherical triangle, (one of those on the com¬ mon base A B, = c) ; and the given surface A + B + C —tt = S. Let I P K be an indefinite perpendicular on the middle of A B, having taken IP = a quadrant, P will be the pole of the arc A B,and the arc CD, drawn through the points PC, will be perpendicular to AB. Let I D ==p.C D = q, the right-angled triangles, A C D, B C D, in which AC =b, BC = a, AD=p + -|c, BD=p — Jc, will give cos a =: cos q, cos (p — c), cos b = cos q cos (p + jtc). But it was found, page 48, that

cot. -i"■, aS = ■——1 4- cos a 4-T cos .b 4-' cos. c sm a sm 67 sin C Substituting in this formula the values cos a -j- cos 6 — 2 cos q cos p cos c, 1 — cos 0=2 cos2 i c, sin b sin 0 = sin c sin B = 2 sin J c cos ^ c sin B; a there results, cot, J, cS = cos—r* i c 4-T cos rp . cos q sin a sin c sin B Again, from the right-angled triangle B C ]}, sin a sin B = sin q; cot A S cos ^ c 4- cos p . cos q sm c sin q or, cos p cos q = cot J S sin c sin q — cos J c. This is the relation between p and q which will determine the locus of all the points C. Produce I P to K, let PK = x. Join KC, andlet KC =y; in the triangle P K C where we have P C = ^ tt — q, the angle KPC = tt — p, the side KC will be found by the formula cos K C = cos KPC sin PK sin P C -|- cos P K cos P C, or cos y = sin q cos x — sin x cos p cos q. Substituting this instead of cos p cos q the value coa | S sin i c sin q — cos | c, there results, cos y = sin x cos c -f- sin q (cos x—sin x cot iV S sin ^ e), SPHERICAL TRIGONOMETKT. 59 in which, if we take cos x — sin a; cot J S sin ^ c = 0, or cot ^ S sin ^ c cot x, there will result cos y = sin x cos J c, and thus a constant value of y is determined. Therefore, if after having drawn the arc IP perpendicular to the middle of the base AB, and beyond the pole the part P K such that cot P K = cot \ S sin ^ c, all the vertices of the triangles on the same base c, and of the same surface S, will be situated on the small circle described from K as a pole at the distance K C, such that cos K C = sin P K cos \ c. This is Lexell's theorem. 48. Given the three sides, B C = a, A 0 = b, A B == c, to find the position of the point I, the pole of the circle cir¬ cumscribing the triangle ABO. Let the angle ACI = #,and the arc AI = CI = BI =0; fn the triangles 0 A I, C B I, we shall have by the equation— cos 0 — cos b cos 0 cos x = :—;—:— sin o sin 0 1 — cos b cot 0 = sin b sin b cot 0, 1 -f cos b . 1 — cos a , cos (0—- 'x) cos (C — x) = cot 0: therefore sin: a cos x (1 -f- cos b) (1 — cos a) or, cos C + sin C tan x — sin a sin b Substituting, in this equation, the values of cos C and sin 0, in terms of the sides a, b, c, and putting for the sake of abridg¬ ment^ M = ^(l—cos2 a—cos2 b—cos2 c-|-2 cos a cos b cos c), . l+cosS—cose —cos a ,. , , we have tan x = — which determines JV1 the angle ACI. From the isosceles triangles ACI, ABI, B 0 I, we have ACI = ^ (C + A — B); and in the same manner, B CI = £ (B + C — A); B AI = £ (A + B — C). From which results these remarkable formulse, tan* i, (A, . +. C — TJXB) ^^1 + COS b — cos a — cos c 60 SPHERICAL TRIGONOMETRY. 1 -j- cos a — cos b — cos c tan J (B + 0 — A) = ^ 1 + cos c — cos a — cos b tan ^ (A + B — 0) = —^^

To which we may add that which gives cot S, and which can he put under the form tani(A„ , 4 + ^B + C) = — 1 — cos a — cos b — cos c

From the value of the tangent of x, already found, we have 1 2 (1 + cos b) (1 — cos c) (1 — cos a) 1 + tan2 x — - c M2 16 cos2 b sin2 J c sin2 J a M2 3 4c cos J b sin -J- c sin J a — ; but, from the equation COS X J\x 1 — cos b cos x = :—=— cot (b = tan i b cot g6, we find sm 6 , tan & b , 4: sin i a sin ^ b sin i c tan 0 = — ; tant 0 = —^- cos x M 2 sin a sin % b sin % c /( . a+6+c . a+b—c . a-f-c—ft- . 64-c —a\ V V8'n 8in^— sm —2~sm 4:9. The surface of a spherical polygon is measured by the sum of the angles, minus the product of two right angles, and the number of sides of the polygon, minus 2. From A draw the arcs AO and AD, the angles of the poly¬ gon ; it will then be divided into as many triangles, minus two, as the figure has sides; but the surface of each triangle is measured by the sum of the angles, minus two right angles; and it is clear, that the sum of all the angles of the triangles is equal to the sum of all the angles of the polygon. i SPHERICAL TKIGONOMETRY. 61 Therefore the surface of the polygon is equal to the sum of the angles diminished by as many times two right angles as the figure has sides, minus two. Thus if S = sum of the angles of a spherical polygon, n = the number of its sides, then the surface of the polygon is S — wtt + 2 7r = S — 2 (n — 2) or S — 2 n -f- 4, when the right angle is taken equal to unity. POLYHEDRONS. 50. If S be the number of solid angles of a polyhedron, H the number of faces, A the number of its edges, then S + H = A + 2. Take a point within the polyhedron, and from which draw lines to all the angular points of the polyhedron; imagine from this point, as a centre, we describe a spherical surface which meets all these lines in as many points, then join these points by arcs of great circles, in such a manner as to form, upon the surface of the sphere, the same number of polygons as there are faces of the polyhedron. Let ABODE be one of these polygons, and n the number of its sides, its surface by the last article will be S — 2 n + 4; S being the sum of the angles A, B, 0, D, E. Similarly if we find the value of each of the other spherical polygons, and add them all together, we conclude that their sum or the surface of the sphere which is represented by 8, is equal to the sum of all the angles of the polygons, less twice the number of their sides, plus four times the number of faces. Now as the sum of all the angles that meet at the same point, A, is equal to four right angles, the sum of all the angles of the polygons is equal to four times the number of solid angles, it is therefore equal to 48. Then double the number of sides AB, BC, CD, &c., is equal to four times the number of edges, or equal to 4 A, since the same edge serves for two faces; 8 = 4S-4A + 4H; or 2 = S—A + H; orS + H = A + 2. [,See Legendre's Geometry, pp. 228, 229."] SPHERICAL TRIGONOMETRY. Got. It follows that the sum of all the plane angles which form the solid angles of a regular polyhedron, is equal to as many times four right angles as there are units in S — 2, S being the number of solid angles of the polyhedron. The plane angles = sum of all the interior angles of each face, which Prop. 32, of the 1st book of Euclid i = H (n — 2). tt = 2 (A — H) tt (since n H = 2 A) •= (S — 2) 2 tt (since A — H = S — 2). 51. There can be only five regular polyhedrons. Since every face has n plane angles, the number of plane angles which compose all the solid angles = wH = Sm = 2Ar and by the last article S + H = A + 2 ;

H = — S, and A = ; substituting these,

S + »S-=S+2, n 2 2nS + 2mS = m?iS -f- 2 n S + 2 m S — m nS = 4/1, S|2(7i + m) — m n} = g _ ^71 2 (n + m) — m n Now this must be a positive whole number, and in order that it may be so 2(m + n) must be greater than mn; and, therefore, — + — > or — > J - —; but m cannot be Tfl 7b 7b TTL less than 3, therefore —ffl cannot be so smallJ as A — 4-3) or 4 consequently, since n must be an integer and cannot be less than 3, it can only be 3, 4, or 5. In the same manner m cannot be less than 3, therefore the values of m can only be 3, 4, or 5. SPHERICAL TRIGONOMETRY. 63 52. To find the inclination of two adjacent a faces of a polyhedron to each other. Let A B be the edge common to the two adjacent faces of a polyhedron, C and E the c centres of the faces. Draw CO, EO, per¬ pendicular to the faces meeting each other in 0; and CD and ED, perpendicular to AB, the intersection of the planes ABC, ABE, then the angle 0DE is the required inclination. Let n be the number of sides in each face, m the number of plane angles in each solid angle if from the* centre 0, and radius equal to unity, describe a spherical triangle meeting the lines OA, 0 0, 0 D in p, q, r, we shall have spherical triangle p q r, in which we have the angle r a right angle, the angle p = and the angle q = — m n cos p and by right-angled triangles, cos qr = —- ; but cos qr = cos COD = sin CDO =sin J C, C being the angle CDE; then TT cos — m sin•in | 0 = TT sm — n This equation is general, and applies successively to the five polyhedrons, by substituting the values of m and n in each case. Tetrahedron m = 3, n =. 3. Hexahedron m — 3, n = 4. Octahedron m = 4, n ass 3. Dodecahedron m = 3, n = 5. Icosahedron m = 5, n = 3. From the triangle pqr from which we have deduced the inclination of the two adjacent faces, we have CO TT TT cos pq~ cot p cot q : or ——- = cot — cot — : 2 OA m n 64 ^ SPHERICAL TRIGONOMETRY. therefore, if we call R the radius of the sphere which cir¬ cumscribes the polyhedron, and r the radius of the sphere inscribed in it, we shall have — tan tan —: r= m n la and, by making the side AB = a, we have 0 A = —^— ; TT sin — n JL a2 and, consequently, R2 = r2 H—-— sin2— n These two equations give for each polyhedron, the values of the radii R and r for the circumscribed and inscribed sphere. We have, supposing 0 known, r = i- a cot — tan \ 0 and R = ir tan — tan ^ 0. 2 n 2 2 m 2 R In the dodecahedron and icosahedron, — has the same r TT TT value for both ; viz., tan — tan —. Therefore, if R be the o O same for both, r will also be the same; that is to say, if these two solids are inscribed in the same sphere, they will also circumscribe the same sphere, and vice versd. The "same property holds with regard to the hexahedron R and octahedron, since the value of — is the same for one as r the other1; viz., tan —n* tanX T-j- •

53. To find the inclination of two adjacent faces in the five regular polyhedrons.* TT cos From the equation sin J 0 = TT, taking the tetrahedron sin — n * Legendre, at page 312, of his Geometry, finds the inclination from cos c — cos a cos b . 0 equation* cos C = : sin:—; a sin b J see equation 3, page 5. SPHERICAL TRIGONOMETRY. 65 where m = o and n = 3, . cos 60° 1 „ sm ^ 0 = ——^5 «= —7=-; /. cos 0 = 4.r sin 60° x/ 3 ' 3 In tlie hexahedron m = 3 and n = 4, sin J 0 = C?S = —7=r, and cos 0 = 0; /. 0 = 90. ^ sm 45 x/ 2 In the octahedron m = 4 and n = 3, . , ^ .cos 45 , / 2 ^ ^ smf 0 = -—ttt = a/ and cos 0 == — 4 ^ sm 60 V 3 * In the dodecahedron m = 3, n == 5, • cos 60 2 1 — y 5 sin %■ G = ——— = — — , and cos O = 7=- sm 36 yio — 2 a/ 5 5 —V 5 In the icosahedron m = 5, n = 3, . , ^ cos 36 l + '/T „ ST sm J C = . g = 7=-, and3 cos C= r— • ^ sm 60A 2 v 5 3

54. To find the solid content ofregular polyhedron.

The area of each face = ^ a2cot-; hence the area of the 4 n Tl TT surface of the polyhedron = H. - a2 cot and the solid con-

area of the surface x by the altitude _ . . tent = = 4 of the area 3 3 of the surface x by radius of inscribed sphere. . n tt nra2 H k = 4- X IITT.- a2 cot — x r = ——— cot— (1) 3 4 n 12 n

^or since r = J a cot ^ . tan 0^

= !i^C0t2^tan^c (2) 66 SPHERICAL TRIGONOMETRY. From either of these equations we can find the solid content. We shall here use the first. Taking the equations

H. tt TT2 cfi2 —/V*r = tan/M — /V*> tan and R — r m n , 2 ^ 4 sm — n we can find r and R.

In the tetrahedron m = 3, n =. 2>; R —T = tan 60 . tan 60 = \/ 3 x 3=3:7 R == 3 r.

(3 }•)2 — r2 = or 8 r2 = y; , a2 a R • • ^ = ht.24:' »' 2?= v/6■ • Also3 — — r

9 3 2^6 In the hexahedron m = 3, n = 4,

a —r = tan 60 tan 45 = v 3 -: r = 2^R 2= .

In the octahedron,m—.4, andn=3, andr=—R——. V 6 v/2 In the dodecahedron, m — 3, n = 5,

r = 1^250 +110-v/s"; R = |(v/15+ \/3). In the icosahedron, m — 5, n = S,

r = ^42+ 18 ST-, R = | \/ioTvV SPHERICAL TRIGONOMETRY. 67 These being substituted in equation (1), page 65, we find the solidity of each of the five polyhedrons. a3 /— For the tetrahedron, the solidity is — v 2;

For the hexahedron a3.

For the octahedron v2. o

For the dodecahedron \/470 + 210 v/5

For the icosahedron — v 14 + 6 v 5.

Examples. 1. Prove that the sine of the sun's declination (S) is a mean proportional between the sines of its altitude at six o'clock (A) and its altitude when due east "(A'). The circles being projected on the meridian, S'M=altitude at six o'clock = A; SC = altitude when due east, cos S'Z = cos Z P . cos P S', or sin A = sin lat. sin b; similarly cos S P = cos SZ cos ZP, or sin 2 = sin A'sin lat.; dividing one by the other we have sin A sin 2 or ———sin S sin A'T.J , sin A : sin S :: sin S : sin A'. 2. Find at what latitude the azimuth of a star, whose de¬ clination = 30°, is equal to 45° at the time of rising. (See last figure.) Let Z be zenith; P the pole; C H horizon; then cos P M = cos P H cos M H, or sin d = sin lat. cos M Z P; but cos M Z P = y=> an

Latitude = 45°. 68 SPHERICAL TRIGONOMETRY. 3. Find the sun's zenith distance when due east on a day whose declination (S) is given, in a given latitude (X). (See fig. p. 67.) Let z c be the prime vertical, P the pole, and S the sun; then cos S P = cos S z cos z P, or sin B = cos S z sin. lat.j , ~ sin $ whence cos S z = — sin X 4. Find latitude of place from given declinations of two stars, and the difference of their altitudes when they are on the prime vertical. Let S and 0 be two stars on the prime vertical; S C = difference of altitudes, and from the three given sides of the A S P 0 we can find the Z. 0, which, with C P in the right- angled triangle zP C, will give us 2 P = 90o-latitude, and the latitude is readily found.

5. Find the relation of the polar to the equatorial diameter of the earth from the horizontal parallax of the moon accu¬ rately observed dt the same time in different latitudes. Let 0 be the latitude, r == corresponding semidiameter of earth, a = equatorial diameter, b = polar distance. Then the equation to the ellipse is a2 y2 + b2 x2 = a2 b2; putting a2 b 2 r cos 0 for x, and r sin 0 for y, r2 = . . , r— • a . a20 sin20 0 + b7<>2 cos2 0

Also, if 0' be another latitude, and / the radius corresponding, 2 2 2 d b dividin one r' = 2 . 2 , 2 2Y7D • S the other, a2 sm2 0 + 672 cos 0' _ _ . a r2 m2 sin2 0' + cos2 0' and making m = -, we have ^ = -2 ^ + and m = /cos2 0'- r'cos^) yf \ t2 sin2 0 — r 2 sin2 0 J

6. To find the augmentation, by parallax, of the moon's diameter. SPHERICAL TRIGONOMETRY. 69 Let M be the moon, o the place of observation on the surface of the earth, whose centre is 0; D = ap¬ parent diameter seen from 0, and D' the apparent diameter seen from 0; then, since the body is the same, D' : D :: CM : oM :: sin z' : sin (z' — p) ; D' = sin z D sin (z' — p) ' sin z' — sin (z' —p) .-.D'-D = D sin (z' —p) sin-^-p cos (z'— ^p) 2D augmentation sought. sin {z! — p) Note.—If A be the value of D' in the last, when the moon is on the horizon, in which case z7 = 90°, and p becomes the horizontal parallax P; D' = D sin— z' r =^ D . sin 90° r = D ; also, we. have Dta = sm (r —p) sin (90 — p) cos r A cos P. 7. From two he¬ liocentric longitudes (AII and AS) and latitudes (PR and QS) to find the place of the node (N) and the inclination of the orbit. We have by Napier's " Circular Parts" (pages 11 and 12), sin NR (or AR—AN) ' sinNS (or AS—AN) _ cotan PNR = tan PR tan QS sin AR cos AN — cos AR sin AN tan PR sin AS cos AN — cos AS sin AN ; hence, tan QS sin AN sin AR .tan QS — sin AS sin PR tan AN = cos A N ' cos A R . tan Q S — cos A S tan P R' 70 SPHERICAL TRIGONOMETRY. AN = required longitude or place of the node; this being found, we have again by Napier's analogies cot P N It (incli- nation offi the orbit)^ =sin AS-AN) • tan Qo 8. Explain Flamstead's method of finding the sun's right ascension. Let b and o be the vernal and autumnal equinoxes. m n When the sun, —-j ] —between b and o, S passing in the ecliptic b m n o, has equal declinations (ms and nv), its distances from the equinoxes (sb and ov) are equal also. We can find when the sun has equal declinations by observing zenith distances for two or three days soon after the vernal equinox, and for two or three days about the same distance of time before the autumnal equinox, and then, by proportion, ascertain the precise time when the sun's declinations are equal; also we can find by proportion the differences of the right ascension of the sun.and some star by observing the differences at noon' for some days. Let E = bs — right ascension of sun after vernal equinox; then 180° — E = b v = right ascension when it has equal declination. A — right ascension of star in former case, and A + P = right ascension of star in latter case. Then we obtain by observation A — E and (180° — E) — (A + P). Let these differences of right as¬ censions be D and D', that is, A — E = D; (180° — E) — (A + P) = D'. Adding these, we have 180° — 2 E — P =

D +D'; E = 18QO~(D + I)/) — P; whence A = D+E Z is known; the value of p, or the change in right ascension of a star between the times of observation, is given by tables. 9. In finding the longitude at sea by the lunar method, to correct the observed distance of the moon from a fixed star from parallax and refraction. This is a useful application of the case 1 of oblique angled spherical triangles. (See page 18.) SPHERICAL TRIGONOMETKY. 71

Let HP be the apparent distance, R being the'Star, and P the moon. As the star's position is elevated by refraction, take IIS = refraction of the star, and take = moon's pa- J \ rallax — refraction, as the parallax being greater than the refraction, the p moon is depressed. Let Z be the zenith; SM will be the true distance, and is thus found. H = apparent altitude of the star, H' = apparent altitude of moon, A = true place of star, A' = that of the moon, A = difference of altitudes, a difference of true altitudes. Then in the triangle Z lip we have cos Hp — cos Z R cos Zp COS • r/ T) • r/ > Sin Z It Bin Zap sin (Z p — Z R) — cos R,p cos A — cos R p 1 — cos Z — sin Z R sin Zp cos H cos H' similarly, sin (Z S — Z M) — cos S M cos a — cos S M 1 — cos Z : ^ sin Z S sin Z M cos h cos h' cos a — cos S M cos A — cos R p 7 T<— = Tv TTT" > an(i we have cos h cos h cos H cos H cos.AcosA' , ^ cos a — cos S M = — — (cos A — cos R p)v ; cos H cos H' ■r-" therefore SM (the true distance) cos A cos h' , cos a — — (cos A — cos R»), cos H cos H' v knowing, by tables, the time when the moon was at this observed distance, so corrected, from a known star, the time it Greenwich is found. 10. To find when that part of the equation of time which arises from the obliquity of the ecliptic is a maximum. PHERICAL TRIGONOMETRY. S = place of the sun; o S = I = longitude; o R = right ascension = r; S R = a = declination; /_ S o R = w = obliquity of ecliptic. Then cos w = cot I tan x (p. 12) tan r 1 — cos w tan v ■•1 + COS U) tan I — tan r sm(l—r) „ , 7—— = ——7; : ; therefore I — r (the equation of tan I + tan r sin (I + r) , time) is a maximum when sin (? + r) is a maximum, i. e. when I + r = 90°. Note.—"When this happens, tan I = cot r, and cot I = tan r, cos w = tan2 r; and cot21 = cos w. 11. In the oblique-angled spherical triangle ABC. Given the side AB 73° 13', the side BO 62° 42', the side AC 119° 5', required the angles. ( A = 44° 18' Ans. B =136° 40' ( 0 = 48° 48' 12. The latitudes and longitudes of three places on the earth's surface, suppose London, Moscow, Constantinople, being given as below : required the latitude and longitude of that place which is equidistant from the former three ? The latitude of London is 51° 30', the latitude and lon¬ gitude of Moscow 55° 45', and 38°, and those of Constan¬ tinople 41° 30' and 29° 15' respectively. 13. Given the latitude of three places, Moscow 55° 30', Vienna 48° 12', Gibraltar 35° 30', all lying directly in the same arc of a great circle. The difference of longitude between Vienna (situated in the middle), and Moscow, easterly, is equal to that between Vienna and Gibraltar, westerly. It is required to find the true bearing and distance of each place from the other, and the difference of longitude, according to the convexity of the globe. 14. Four given equal spheres being placed in close contact with each other, it is required to find the volume of the space inclosed between them and the three triangular planes through each three centres. SPHERICAL TRIGONOMETRY. 73 15. A point P being taken in the surface of a sphere, let a, /3 denote its spherical distances from two given points ; then if m cos a + n cos = a constant quantity, m and n being any given numbers, the locus of P will be a circle. 16. The base of a spherical triangle is given, and the sum of the cosines of the angles at the base, to trace the locus of its vertex. 17. The sides of a spherical triangle are produced to meet again in three more points, thus forming, with the original, four spherical triangles, which constitute Davies's "Associated Triangles" (12th edit., Hutton's Course, vol.ii. p.41), rr^rz are the radii of the inscribed, and R Ri R2 Rs the radii of the circumscribed circles. Prove that tan2 R + tan2~Ri + tan2 R2 + tan2 R3 = cot2 r + cot2 ri + cot2 T2 + cot2 7*3. 18. A person engages to travel from London to Constan¬ tinople, and to touch the equator in his journey; required the point of contact and the length of his track, admitting it to be the shortest possible, and the earth a sphere. 19. The angular points of two triangular pyramids being respectively situated on four converging lines in space, let the corresponding faces be produced to meet; then will the four lines of section be all situated in the same plane. 20. Given the longitudes- of two places, 6° 49', and 54° 00', their respective latitudes 48° 23' 14", and 4° 56' 13'; find their distance^ the longitudes being both west, and their lati¬ tudes both north.

INDEX.

Adams's method of reducing Napier's Napier's circular parts, described, 11; analogies, 9. rules, 11; examples, 12; application to Analogies, Napier's formulae, 8; Adams's an ambiguous case, 15. method of reducing, 9. Angle, to reduce an, to the horizon, 26. Oblique-angled spherical triangles, solu¬ Angles and sides of spherical triangles, tion of, 18, 35:—(1) given the sides, fo relations between, 5. find three angles, 18, 35; (2) given the two sides, and the angle opposite to one Circle, arc of a, defined, 1. of them, to find the remainder, 19, 36; Circle, great, of a sphere, defined, 1; (3) given the two sides, and the in¬ poles of a great circle, 1. cluded angle, to find the remaining Circular parts, Napier's, described, H; angles and side, 20, 37; (4) given the application to an ambiguous case, 15; two angles, and the adjacent side, to rules, 11; examples, 12. find the remaining angle and sides, 20, Cosines, to determine the cos and sines 39; (5) given two angles, and the side of a spherical triangle, in terms of the opposite to one of them, to find the rest, cosines and sines of the sides, 3. 21, 41; (6) given the three angles, to find the three sides, 22, 42. Definitions, 1. Octahedron, to find the solid content, 66; Dodecahedron, to find the solid content, solid content, 67. 66; solid content, 67. Parallelepiped, given the three edges and Examples and problems, 67. angles between them, to find the solidity, 55; the same being given, to find the Fundamental formulae, 4. diagonal, 56. Polar triangle described, 2. Girard's theorem for the areas of spherical Polygon, spherical, to find the area, 60. triangles, 51. Polyhedrons, to find the inclination of two Great circle, definition, 1. adjacent faces in the five regular, 64. Polyhedrons, solid content of regular, 65, Hexahedron, to find the solid content, 66; 67; equations to, 63; to find the area solid content, 67. of formulae, 61; to find the inclination Horizon, to reduce an angle to the, 26. of two adjacent faces to each other, 63. Problems and examples, 67. Icosahedron, to find the solid content, 66; solid content, 67 Quadrantal triangle, 3. Isosceles triangle, 3. Quadrantal triangles, solution of, 33. Legendre's theorem, for the solution of Eight-angled spherical triangles, 3; nu¬ spherical triangles whose sides are very merical solution of, 28. small compared with the radius of the sphere, 52. Sides and angles of spherical triangles, Lexell's theorem, explained, 57. relations between the, 5. Lhuillier's formula for the area of a sphe¬ Sines, to determine the sines and cosines rical triangle, 50. of a spherical triangle, in terms of the sines and cosines of the sides, 3. Napier's analogies: formulae, 8; Adams's Solid, content of the five polyhedrons, to method of reducing, 9. find the, of the dodecahedron, 66; of 76 INDEX. the hexahedron, 66; of the icosahe- one of the sides and the three angles, * dron, 66; of the octahedron, 66; of the ambiguous cases, 23; properties of, 27; tetrahedron, 66. oblique angled, solution of, 18, 35, see Sphere, definition of the, 1; area of the also Oblique-angled triangles; right- lune, is proportional to the angle, 44; angled, solution of, 10; right-angled, circumference, ratio of, to the surface, numerical solution of, 28:—(1) when 44; every section of a sphere is acircle^ the hypotlienuse and one side are given, 1; great circle of the sphere defined, 1; 28; (2) when two sides are given, 30; to determine a line on the surface of a, (3) when a side and its opposite angle on which the vertices of all the trian¬ are given, 31; (4) when a side and its gles are situated. 57. adjacent angle are given, 32. Spherical excess, defined, 45 ; formula Spherical trigonometry described. 3; fun¬ for, 46; remarks on, 45. damental formula, 4. Spherical polygon, to find the area, 60. Supplemental triangle, 2; sum of a, 2. Spherical triangle, defined, 1; angles of, 1; area of, described, 44; to find the Table of results from ambiguous cases. 23. area in terms of the three sides, 48; Tetrahedron, to find the solid content of Girard's theorem, 51; Lhuillier's for¬ a, 66 ; solid content, 67. mula, 50; Legendre's theorem for the Triangle, polar, 2; sum of a, 2. solution of, whose sides are very small Triangles, spherical, defined, L; angles of, compared with the radius of the sphere, 1; described, 3; to determine the sines 52; sum of a, 2; to determine the sines and cosines of, 3; isosceles, 3; qua- and cosines of a, in terms of the sines drantal, 3; right-angled, 3; ambiguous and cosines of the sides, 3; relation be¬ cases, 23; Girard's theorem for the tween the sides and angles of, 5:—(1) areas of, 51; Legendre's theorem for relation between three sides and an the solution of, 52; oblique-angled angle, 5 ; (2) relation between two spherical, solution of, 18, 35; quad- sides and their opposite angles, 5; (3) rantal, solution of, 33, see also Spherical relation between two sides and their triangles. included angle, and the angle opposite Trigonometry, spherical, described, 3; one of them, 6; (4) relation between fundamental formula, 4.

THE END.

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