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Originally appearing in Volume V27, Page 277 of the 1911 Encyclopedia Britannica.
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DCB and draw FG perpendicular to BC, then CG=I(a—b,e FBE=1(A+B), LFCG=9o°—IC. From the triangle CFG we have See also:

cos CFG=cos CG See also:sin FCG, and B from the triangle FEB cos EFB = cos EB sin FBE. Now the angles g CFG, EFB are each supplementary cosl (a —b)cosl C =sinl (A +B)cosic. (13) Also sin CG=sin CF sin CFG and sin EB=sin BF sin EFB; therefore sinl(a—b)cosiC=sinl(A —B)sinlc. (14) Apply the formulae (13), (i4) to the associated triangle of which a, jr—c, A, lr—B, a-C are the sides and angles, we then have sing(a+b)sinZC=cosl(A —B)sinic (15) cosl(a+b)sinlC=cosl(A+B)coslc. (16) The four formulae (13), (14), (15) (16) were first given by See also:Delambre in the Connaissance See also:des Temps for 1808. Formulae See also:equivalent to these were given by Mollweide in See also:Zach's Monatliche Correspondenz for See also:November i8o8. They were also given by See also:Gauss (Theoria motus, 1809), and are usually called after him. 11. From the same figure we have See also:Napier's tan FG=tan FCG sin CG=tan FBG sin BG; Analogies. therefore cotlCsinl(a—b)tanl(A —B)sinl,(a+b.), or tanl(A —B) =s!n 2(a+b)cotlC. (17) Apply this formulae to the associated triangle (a—a, b, -ir—c, it—A, B, it — C), and we have cot l(A+B)=cos 1(a+b)tan IC, cos z(a—b) or tan 5(A +B) (a+b)cot lc. (18) If we apply these formulae (17), (18) to the polar triangle, we have sin RA —B tan I(a—b) =sin 1(A+B)tan lc tan 1(A+B)=cos (A—B)tan ic.

cos RA +B) cos la cos b sin C = cos c cos l (A +B — C) sin la sin b sin C = cos c cos 1(A +B +C) These formulae were given by Schmiesser in Crelle's Journ., vol. x. The relation sin b sin c+cos b cos c cos A =sin B sin C—cos B cos C cos a was given by Cagnoli in his See also:

Trigonometry (1786), Cagnolfs and was rediscovered by See also:Cayley (Phil. Msg., 1859). Formulae. It follows from (I), (2) and (3) thus: the right-See also:hand See also:side of the See also:equation equals sin B sin C+cos a (cos A - sin B sin C cos a) =sin B sin C sin' a+cos a cos A, and this is equal to sin b sin c + cos A (cos a—sin b sin t cos A) or sin b sin c + cos cos a cos A.. 13. The formulae we have given are sufficient to determine three parts of a triangle when the other three parts are given ; moreover such formulae may always be chosen as are adapted See also:solution of to logarithmic calculation. The solutions will be unique Triangle except in the two cases (1) where two sides and the See also:angle opposite one of them are the given parts, and (2) where two angles and the side opposite one of them are given. Suppose a, b, A are the given parts. We determine B from the See also:formula sin B=sin b sin A/sin a; this gives two Ambiguous supplementary values of B, one acute and the other Gases. obtuse. Then C and c are determined from the equations sin I(A+B tan lC=sin 1(a+b) cot l(A —B), tan 1c =sin (A —B) tan 1(a—b). sin Now tan IC, tan lc, must both be See also:positive; hence A—B and a—b must have the same sign.

We shall distinguish three cases. First, suppose sin b<sin a; then we have sin B<sin A. Hence A lies between the two values of B, and therefore only one of these values is admissible, the acute or the obtuse value according as a is greater or less than b; there is therefore in this See also:

case always one solution. Secondly, if sin b> sin a, there is no solution when sin b sin A > sin a ; but if sin b sin A<sin a there are two values of B, both greater or both less than A. If a is acute, a—b, and therefore A —B, is negative; hence there are two solutions if A is acute and none if A is obtuse. These two solutions fall together if sin b sin A =sin a. If a is obtuse there is no solution unless A is obtuse, and in that case there are two, which coincide as before if sin b sin A =sin a. Hence in this case there are two solutions if sin b sin A<sin a and the two parts A, a are both acute or both obtuse, these being coincident in case sin b,sin A =sin a; and there is no solution if one of the two A, a, is acute and the other obtuse, or if sin b sin A> sin a. Thirdly, if sin b=sin a then B=A or it =A. If a is acute, a-b is zero or negative, hence A —B is zero or negative; thus there is no solution unless A is acute, and then there is one. Similarly, if a is obtuse, A must be so too in See also:order that there may be a solution. If a =b = .ir, there is no solution unless A See also:fir, and then there are an See also:infinite number of solutions, since the values of C and c become indeterminate.

The other case of See also:

ambiguity may be discussed in a similar manner, or the different cases may be deduced from the above by the use of the polar triangle transformation. The method of See also:classification according to the three cases sin b-sin a was given by See also:Professor See also:Lloyd See also:Tanner (Messenger of Math., vol. xiv.). 14. If r is the angular See also:radius of the small circle inscribed in the triangle See also:ABC, we have at once tan r=tan IA sin (s — a), where 2s=a+b+c; from this we can derive the formulae tan r=n cosec s= 1N sec IA sec IB sec IC= Radii of sin a sin IB sin IC sec IA (21) Relates to where n, N denote the expressions {sins sin (s—a) sin (s—b) sin (s—c)}1, Triangles. { —cos S cos (S—A) cos (S—B) cos (S—C){l. The escribed circles are the small circles inscribed in three of the associated triangles; thus, applying the above formulae to the triangle (a, sr—b, it—c, A, a—B, it — C), we have for ri, the radius of the escribed circle opposite to the angle A, the following formulae tan ri=tan IA sin s=n cosec (s—a) =lN sec IA cosec IB cosec IC =sin a cos IB cos IC sec A. (22) The See also:pole of the circle circumscribing a triangle is that of the circle inscribed in the polar triangle, and the radii of the two circles are complementary; hence, if R be the radius of the circumscribed circle of the triangle, and Ri, See also:R2, R the radii of the circles circumscribing the associated triangles, we have by See also:writing iir—R for r, fir—Ri for ri, a—a for A, &c., in the above formulae cot R =cot Tacos (S—A)=ln cosec la cosec b cosec lc =—N sec S =sin A cos lb cos lc cosec la (23) cot Ri= — cot la cos S= In cosec la sec lb sec lc =N sec (S—A) =sin A sin lb sin lc cosec la. (24) The following relations follow from the formulae just given: 2tanR =cot ri+cot r2+cot r3 —cot r, 2tanRi =cot r +cot r2 +cot r3 —cot tan r tan ri tan r2 tan ra =n', sin' s =cot r tan ri tan r2 tan ra, sin' (s—a) =tan r cot ri tan r2 tan r2. 15. If E=A+B+C—r, it may be shown that E Formulae multiplied by the square of the radius is the See also:area of for the triangle. We give some of the more important Spherteai expressions for the quantity E, which is called the Excess. spherical excess. We have cos l(A + B) cos 2(a + b) sin 1(A it B) — cos l(¢ — b) sin lC cos is and cos iC cos lc sin 1(C — E) cos 1(a {- b) cos EC — E) _ cos l(a — b), or sin iC cos lc and cos IC — cos lc sin -C — sinl(C—Ecos lc— cosl(a+b). hence sin IC + sin 1(C — E) = cos lc + cos l(a + b)' D (19) (20) The formulae (17), (18), (19), (20) are called Napier's " Analogies "; they were given in the Mirif. See also:logar. canonis descriptio.

i2. If we use the values of sin la, sin 1b, sin lc, cos la, cos lb, cos lc, given by (9), (Io) and the analogous formulae obtained by interchanging the letters we obtain by multiplication Schmelsser's sin la cos b sin C=sin lc cos 1(B+C—A) Formulae. therefore tan }E The theory of the trigonometrical functions is intimately connected ; (s—c). with that of complex numbers—that is, of See also:

numbers of the See also:form tang* (C—E) =tan is tan Similarly tan IF tanz -,'(C—E)=tan j;(s—a) tan 2(s—b); (25) x+iy(t = s/ —I). Suppose we multiply together, by the connexion therefore *E=(tan Is tan 2(s—a) tan 1(s—b) tan 1(s—c)}2 rules of See also:ordinary See also:algebra, two such numbers we have with Theory tan (xi + Lyi) (x2 + ty2) = (xix2—yiy2) + t(xiyz + xzyl). of Complex We observe that the real See also:part and the real See also:factor of the Quantities. imaginary part of the expression on the right-hand side of this equation are similar in form to the expressions which occur in the addition formulae for the cosine and sine of the sum of two angles; in fact, if we put xi = Ti cos 01, yi = ri sin 01, x2 = r2 cos 02, y2 = r2 sin 02, the above equations becomes This formula was given by J. Lhuilier. Also cos i 1(a+b) sin IC cos 1E—cos IC sin ZE sin IC; cos lc cos 2(a 1E= b) C; cos IC cos ZE+sin IC sin cos cos ~c whence, solving for cos 1E, we get +cos a+cos b +cos c cos 2E= 4 cos la cos 2b cos lc This formula was given by See also:Euler (Nova acra, vol. x.). If we sin 1E from this formula, we obtain after reduction 2 cos la cos lb cos lc a formula given by Lexell (Acta Petrop., 1782). From the equations (21), (22), (23), (24) we obtain the following formulae for the spherical excess: sin"-4E=tan R cot R1 cot R2 cot R2 4(COt rl+cot r2+cot r3) =(cot r—cot ri+cot r2+cot 7'3) (cot r+cot r1—cot r2+cot r3)X (cot r+cot ri+cot r2+cot r3). The formula (26) may be expressed geometrically. Let M, N be the See also:middle points of the sides AB, AC.

Then we find cos MN 1 +cos a+cos b+cos c; hence cos E =cos MN sec i2a. 4 cos lb cos 2c z z A geometrical construction has been given for E by Gudermann (in Crelle's Journ., vi. and viii.). It has been shown by See also:

Cornelius Keogh that the See also:volume of the parallelepiped of which the radii of the See also:sphere passing through the middle points of the sides of the triangle are edges is sin 1 E. Proropertied 16. Let See also:ABCD be a spherical See also:quadrilateral inscribed P in a small circle; let a, b, c, d denote the sides AB, BC, SPherical CD, DA respectively, and x, y the diagonals AC, BD. lateral It can easi_y be shown by joining the angular points inscribed of the quadrilateral to the pole of the circle that In small A+C=B+D.

End of Article: DCB

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