Page:EB1911 - Volume 27.djvu/291

Rh By division we have

A sin $(a+c-b) sin %(a+b-c) fi

tan; S sin § (b-|-c-a) sin § (a-I-b-l-c) (7) and by multiplication

sinA =2{sin (a-~b+c) sin § (b+c-a)sin § (c-l-a-b) sin § >(a+b-c)}§ sin b sin C=lI —cosfa-coszb-coszc +2 cos a cos b cos c}5 sin b sin 0. Hence the quantity k in (3) is

{1 -cos'a-cos' b-COS2C+2 cos a cos b cos c}l/sin a sin b sin c. (8) 0fHalf- Apply the polar triangle transformation to the formulae sides. (5), (6), (7) (82'a.nd we obtaiin B C a    A

cos '(A-|- B) cos (+

COS2 i 2 sin B sin cf i (9)

a -cos l(B-l-C-A) cos l(A+B~|-C lf

S'“5= 2 sin B Sin cz 5 (10)

a -cos %(B-l-C-A) cos § (A +B+C i H) WE" COS;<A+c-B) ws;<A +B~c

lfk' ={1 -cos?/I -cos2B-cos2C-2cosA cosBcosC}%/sinA sinBsinC, we have. k/€'=I (12)

10. Let E be the middle point of AB; draw ED at right angles to D ' AB to meet AC in D; then DE

Del b, bisects the angle ADB.

C Fm;ZI;:s Let CF bisect the angle

° DCB and draw FG perpendicular

to BC, then

CG=%(a-b, L FBE=§ (A--B),

LFCG=9O°-%C.

From the triangle CFG we have

cos CFG=cos CG sin FCG, and

from the triangle FEB cos EFB=

A Bcos EB sin FBE. Now the angles

3 CFG, EFB are each supplementary

FIG. 7. to the angle DFB, therefore cos§ (a - b)cos%C = siné (A -I-B)cos%c. (13) Also sin CG=sin CF sin CFG and sin EB=sin BF sin EFB; therefore sin§ (a-b)cos§ C=sin%(A -B)sin§ c. (14) Apply the formulae (13), (14) to the associated triangle of which a, 1r-b, 1r-c, A, 1r-B, vr-C are the sides and angles, we then have

sin§ (a+b)sin%C==cos§ (A -B)sin-$0 (15) cos%(a+b)sin§ ;C=cos§ (A -{-B)cos§ c. (16) The four formulae (13), (14), (15) (16) werehrst given by Delambre in the Connaissance des Temps for 1808. Formulae equivalent to these were given by Mollweide in Zach's Monatliche Correspondenz 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 S lug f to logarithmic calculation. The solutions will be unique T; 01:0 except in the two cases (1) where two sides and the angle aug 5° 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 supplementary values of B, one acute and the other ca obtuse. Then C and c are determined from the ses equations

Ambiguous

1 . 1

"ffl-bl ~(A-ltan

§ C= cot %(A -B), tan §  tan § (a-b). I

Now tan § C, tan ic, 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; 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 case always one solution. Secondly, if sin b>sin a, there is no solution when sin b sin A > sin at; 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 bgng 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 bsin A>sin a. Thirdly, if sin b=sin a then B=A or -/r=A. If a is acute, 11-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 order that there may be a solution. If a=b=%1r, there is no solution unless A =%-/r, 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 bgsin a was given by Professor Lloyd Tanner (Messenger of Math., vol. xiv.). 14. If 1 is the angular radius of the small circle inscribed in the triangle ABC, we have at once tan 1=tan 5A sin (J-2s=a.+b+c; from this we can derive the formulae a), where

tan 1=1L cosec s=§ N sec § A sec %B sec %C= Refill “f sin a sin § B sin § C sec 'EA (21) cimles where 11, N denote the expressions ';7"t°'} t° rang es.

{sinssin (s-a) sin (s-b) sin (s-c)}%, {-cos S cos (S-A) cos (S-B) cos (S-C)}§. Formulae. 2

for November 1808. They were also given by Gauss (Theoria moms, 1809), and are usually called after him. 11. From the same figure we have

Napiefs tan FG=tan FCG sin CG=tan FBG sin BG; Analogies. therefore cot4;Csin§ (a-b)tan§ (A -B)sin%, (a-Hz), 1 b

or tan§ (A-B) = cotéC. (17)

Apply this formulae to the associated triangle (vr-a, b, 1r-c, -/r-A, B, 1I'“C), and we have

¥ b

cot %(A -l-B) = § C,

1 b

or fan;(A -1-B) = ac. (is)

If we a l these formulae (17), (18) to the polar triangle, we have PP Y

1 =  1

tan, (a b) Sin é(/1+B)tan, c (19)

L.

nm l(A+B)=9 @tan lc. (20)

2 cos § (A -1-B) 2

The formulae (17), (18), (19), (20) are called Napier's “ Analogies ”; they were given in the Mirzf. logar. canonis de.rc1iptio. 12. If we use the values of sin $11, sin éb, sin éc, cos éa, cos éb, cos § c, given by (2), (10) and the analogous formulae obtained by interc anging the letters we obtain by multiplication 5°h"'°'“"°'3 sin éa cos éb sin C=sin lc cos § (B-l-C-A) The escribed circles are the small circles inscribed in three of the associated triangles; thus, applying the above formulae to the triangle (a,1r-b, -ir-c, A, 1r-B,1r-C), we have for 11, the radius of the escribed circle opposite to the angle A, the following formulae tan 1, =tan § A sin s=n cosec (s-a) =§ Nsec 5/l cosec § B cosec § C =sin a cos § B cos %C 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 Rl, RQ, R the radii of the circles circumscribing the associated triangles, we have by writing évr-R for 1, in-R1 for 11, 1r-a for A, &c., in the above formulae cot R=cot be cos (S-A) =§ n cosec éa cosec ib cosec § c=-N sec S =sin A cos ib cos sc cosec ia (23)

cot R1= - cot ia cos S=%1'l cosec sa sec $11 sec éc=N sec (S-A) =sin A sin éb sin éc cosec éa. (24) The following relations follow from the formulae just given:- 2tanR =cot 11-l-cot 12-I-cot 13-cot 1, 2tanR1 =cot 1 +cot 124-cot 13-cot 11, tan r tan 11 tan 12 tan 13 =n2, sinz .S =cot 1 tan 11 tan 12 tan 13, sin2 (s-U.) =tan r cot 1, tan 12 tan 13. 15. If E =A +B'l'C*1F, it may be shown that E cos éa cosébsin C=coséccos§ (A -l-B-C). sin ia sin éb sin C=cos iccos § (A +B~}-C) These formulae were given by Schmiesser in C1elle's Joum., vol. x. The relation sin b sin c-l-cos b cos c cos A=sin B sin Coos B cos C cos a was given by Cagnoli in his Trigonomelry (1786), ca Mrs and was rediscovered by Cayley (Phil. Mag., 1859) Ffluhm It follows from (1), (2) and (3) thus: the right-han ° side of the equation equals sin B sin C -1-cos a (cos Asin B sin C cos a) =sin B sin C sin* a-l-cos a cos A, and this is equal to sin b sin c -l-cos A (cos a-sin b sin c cos A) or sin b sin c + cos b cos c cos A.

ti

multiplied by the square of the radius is the area of £°'m"l'” the triangle. Ne give some of the more important Srherkal the quantity E, which is called the' E§ cess We have

cos %'(A -i- B): cos § (a + b) ard sin § (A + B) = cos *Ha - b), sin 5C cos ie ' cos éC cos éc

sin § (C - E) Cos é(a il- b) cos § (C - E) cos § (a - b) or sin %C = T cos és  and cos § C " cos éc he sin § C - sin § (C -   = cos $6 - cos § (a -1- b) nee sm °}C + sm %(C - E) cos éc + cos § (a + b);