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 cos CFG=cos CG 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, aC 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 Delambre in the Connaissance des Temps for 1808. Formulae equivalent to
these were given by Mollweide in Zach's Monatliche Correspondenz for November i8o8. They were also given by Gauss (Theoria motus, 1809), and are usually called after him.
11. From the same figure we have
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 Trigonometry (1786), Cagnolfs and was rediscovered by Cayley (Phil. Msg., 1859). Formulae. It follows from (I), (2) and (3) thus: the righthand
side of the 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 solution of to logarithmic calculation. The solutions will be unique Triangle except in the two cases (1) where two sides and the 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 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 positive; hence A—B and a—b must have the same sign. We shall distinguish three cases. First, suppose sin b sin a, there is no solution when sin b sin A > sin a ; but if sin b sin A sin a. Thirdly, if sin b=sin a then B=A or it =A. If a is acute, ab 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 order that there may be a solution. If a =b = .ir, there is no solution unless A fir, and then there are an infinite number of solutions, since the values of C and c become indeterminate.
The other case of 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 classification
according to the three cases sin bsin a was given by Professor Lloyd Tanner (Messenger of Math., vol. xiv.).
14. If r is the angular radius of the small circle inscribed in the triangle 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 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, R2, R the radii of the circles circumscribing the associated triangles, we have by 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 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. 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 numbers of the 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 ordinary 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 part and the real factor of the Quantities.
imaginary part of the expression on the righthand 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 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 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 Cornelius Keogh that the volume of the parallelepiped of which the radii of the sphere passing through the middle points of the sides of the triangle are edges is sin 1 E.
Proropertied 16. Let ABCD be a spherical 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. If we use the last expression in (23)
Ctrcle, for the radii of the circles circumscribing the triangles
End of Article: DCB
