THE ELEMENTS OF SPHERICAL TRIGONOMETRY. BY JAMES HANN, J,\\TE MATHEMATICAL ...

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 1 a plane passing through it, draw 00 to the cutting plane, and pro¬ N M duce it both ways to D and E, and V draw the radii of the sphere 0 A, OP. ' Now, since OOP and OCA are right anales, O A2 2 — 0 C22 = A C2. 2 2 and OP -OC = PC , but 0A = 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 b2

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 7 = cos 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-circumj.b ferences C A 0', 0 B C, and make the ... - triangle ABC of which the sidea v a' and b', or B C and A C, are supc7 •*'" 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, / ; intersect in A', which forms the c ./ /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 By applying equation (1) to the three angles, we cos a = cos b cos c + sin b sin c cos A cos b — cos a cos c + sin a sin c cos B cos c = cos a cos b -f sin a sin b cos C

angle. have (1) (2) (3)

13. 2nd, Relation between two sides and their opposite angles. From equation (1) we have . cos a — cos b cos c cos A = ;—7—;— sin o sin c tt . 2. A a —« ,cos. b» cos ~ c)s A = -1 — cos22A Hence sin A = .1 — (cos . sin'4 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 a

sin B sin b

sin 0 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 cot b sin a = cos a cos 0 + sin 0 cot B cot a sin c = cos c cos B + sin B cot A cot c sin a = cos a cos B + sin B cot 0 cot b sin c = cos c cos A + sin A cot B cot c sin b = cos b cos A 4- sin A cot 0

(5) (6) (7) (8) (9) (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 ' cos b — cos a cos c sin A cos B we have 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) / , . sin a sin A ana since-:—-—sin b sin B

(a)

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)'

we have

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) 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) 5 sin b sin c sin IL A = . /Bi'1 i (« + b — c) sin j (a — 6 + c) V sin b sin c For the sake of abridgment, put a + 6 + c=:2s, and the preceding expression becomes /a ( n( c) »A=v 'T\o)° V "sin sin c'- . me way In the same /Bi'iSBin (s-q). cos iA== 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 i2 V(a — b); Bin^(A + E) ^ ein£(A — B) 6in f (« + &), 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

cot i 0 = tan i (A — B) .

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 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) V sin B sm 0 sin S sin (A — S)7 >.-' tan J a = a // -^-7^ gx \ /n Q V sin (B — S) sm (0 — o) 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 -hJ a = x / ^ ; =r^ ^ V sm B sm 0 - /— cos S . cos (S — A) tan & a ios (S _ B) cos (SC) A

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 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 . B -D = sin b; sin A sin ; 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. two solutions, a180 two solutions, a + 51800 no solution. a%b two solutions, ) = 90o| no solution. a>6 a + 6 > 180° b 90o-

&>90o & = 90(

a

^Cb a-\-b :i8oo ja>5

two solutions, one solution, no solution. two solutions, one solution, no solution. two solutions, no solution.

one solution, no solution, no solution. one solution, A == 90° \ no solution, b >90° I a + 6 = 90° | a90° no solution. b 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' tan b1 = tan a' cos C

sin c' = sin a1 sin C; tan cf = tan a' cos B' o3

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 We sin B sin V 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. AB = 79° 17' M" \ BO = 110° (to find the rest. AC =r 58° j To find the angle A. sin - 6 ) sil1 ("-"). V 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

By page 18, sin J A =

A/

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. tt , . sin 6 Here by page• 6, sm Br, = sin A; 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 J2 O = tan J v(A n + B)J CQS % (a + 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 log cos -J- (a — 6) = log cos 3° 24/ 25" = log cot J 0 = ... jo = 82° 30' 39"; or O = 165° 1' 18".

19-1180197 9*9992318 9*1187879

We might find c from the equation sin C sin c = sm a . -—r, sm A 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 J2 (A = cot}3 C . 003 f ~ ^ K + B)' cos i (a + 6) and tan i = cot ^ 0 . 8laf{a~^ 2 (A v — B) ^ 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" J (a—b) = 20 0 22

log cot = 104785395 log sin = 9*534:1789

J (a + 6) = 64: 14:

20-0127184: log sin = 9-9545255

7

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 5t sin a . sin o .\ 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 = ver-

. ,

/ sin a . sin 6 . ver. sin 0 \ \ 1 + ver. sin (a—b) } sin a . sin b . ver. sin 0 ver. sin (a—6)

8,n

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 o — 6 = 40

0 44

.-. 2 log tan 0 and log tan 0

29-1398419 ver sin = 9*3693878 = 19-7704541 = 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^ c . 008 i 5} T 6) ^ tan 2 ' cos£(A + B) tan £ (a - b) = tan £2 c . S|n f (A ~ B) ^ ' 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 log cos ^ (A + B) = log cos 82° 14' log tan J (a -f b)

19-4973920 = 9-1307812 = 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 log sin £ (A + B) = log sin 82° 14'

19-5390185 = 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 a . « . or, sin U = sin c

sin G sin c sin A . ~— 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 log sin a = log sin 85° 59' Jog sin 0

19-7693370 = 9*9989319

= 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: = 9*974:7475

log sin A = log sin 70° 39'

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'/.

t

To find c. By Napier's Analogies tan

003 c = tan £(« 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 A B mnA)% then each of the angles at 0 and e 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 * + tan and R2 — r2 r /V* /M m n , ^ 4 sm2 — n we can find r and R. In the tetrahedron m = 3, n =. 2>; R — = tan 60 . tan 60 = \/ 3 x T

(3 }•)2 — r2 =

3=3:7

R == 3 r.

or 8 r2 = y;

, a2 a R • • ^ = ht. • Also ——r 24:' »' 2?=■ v/6 3 9

3

2^6 In the hexahedron m = 3, n = 4, — = tan 60 tan 45 = r

v

3 -: r = 2 ^R 2 = a

.

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; a3.

For the hexahedron For the octahedron For the dodecahedron For the icosahedron

o

v2.

\/470 + 210 v/5 —

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 ———T.J , sin A : sin S :: sin S sin A' 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< s n

l i ^

Latitude = 45°.

; .'.•£■ = sin lat. —, and sin lat. =

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 Po in the rightangled triangle zP C, will give us 2 P = 90 -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 2 r cos 0 for x, and r sin 0 for y, r— • a r = 20 . 20 . , 72 . a sin 0 + b cos2 0 Also, if 0' be another latitude, and / the radius corresponding, r'2 =

d2 b2 . 2 , Y7D • a sm 0 + 6722 cos 2 0' 2

dividin

S one

the other,

_ _. 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) ' .-.D'-D = D

sin z' — sin (z' —p) sin (z' —p)

sin-^-p cos (z'— ^p) 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; z'r =^ , we.have D ta = D' = D sin — D . sin 90° r = D ; also 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' 2D

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^ = sin AS-AN) • nation offi the orbit) 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 = + 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.

D +D'; E =

18QO

~(D

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.)

71

SPHERICAL TRIGONOMETKY.

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 paJ \ 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 1 — cos Z —

sin (Z p — Z R) — cos R,p cos A — cos R p sin Z R sin Zp cos H cos H'

similarly, 1 — cos Z : ^

sin (Z S — Z M) — cos S M cos a — cos S M sin Z S sin Z M cos h cos h'

cos a — cos S M cos A — cos R p 7 T