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Encyclopedia Britannica - Main :: NUM-ORC |
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OBC . Then the projection of OA on OB is the sum of the projections of OL, LM, MA on the same straight line. Since AM has no projection on any straight line in the Arndt,. plane OBC, this gives angles. mental OA cos c = OL cos a+LM sin a. Equations Now OL=OA cos b, LM=AL cos C=OA sin b cos C; between therefore cos c =cos a cos b+sin a sin b cos C. Sides and We may obtain similar formulae by interchanging the Angles. letters a, b, c, thus cos a =cos b cos c+sin b sin c cos A cos b=cos c cos a+sin c sin a cos B cos c=cos a cos b+sin a sin b cos C These formulae (I) may be regarded as the fundamental equations connecting the sides and angles of a spherical triangle; all the other relations which we shall give below may be deduced analytically from them ; we shall, however, in most cases give independent proofs. By using the polar triangle transformation we have the formulae cos A = cos B cos C+sin B sin C cos a cos B = cos C cos A +sin C sin A cos b (2) cos C= cos A cos B+sin A sin B cos c In the figures we have AM=AL sin C=r sin b sin C, where r denotes the radius
therefore sin B sin c =sin C sin b. By interchanging the sides we have the equation sin A sin B sin C_k sin a sin b sin c we shall find below a symmetrical form for k. If we eliminate cos b between the first two formulae of (I) we have cos a sin''-c=sin b sin c cos A +sin c cos c sin a cos B; therefore cot a sin c = (sin b/sin a) cos A +cos c cos B =sin B cot A +cos c cos B. We thus have the six equations cot a sin b=cot A sin C+cos b cos C-cot b sin a=cot B sin C+cos a cos C cot b sin c =cot B sin A+cos c cos A cot c sin b =cot C sin A+cos b cos A cot c sin a-=-cot C sin B+cos a cos B cot a sin c=cot A sin B+cos c cos B- When C=2,r formula
and (3) gives from (4) we get The formulae and follow at once from (a), (3), (y). These are the formulae which are used for the solution of right - angled triangles. Napier gave mnemonical rules for remembering them. The following proposition follows easily from the theorem in equation (3): If AD, BE, CF are three arcs drawn
the angles to the middle points of the opposite sides, each pass through a point. 9. If D be the point of intersection of the three Formuim bisectors of the angles A, B, C, and if DE be drawn
Halt
sin c sin ADB sin b sin ADC therefore sinz IA also sin BD = sin IA sin CD = sin 2A , sin BD sin CD sin CDE sin BDE But sin BD sin BDE =sin BE sin b sin c =sin z(a+cb), and sin CD sin CDE=sin CE=sin 2(a+bc); therefore sA csin 2(a+cb) sin 1(a+bc) z 2 sin b sin c (5) in Apply this formula
A )sin z(b+ca) sin 1(a.+b+c) ; (6) - = the formula cos sin b sin c ) (2) The two sides a, b and the included angle
A+B=rC, L tan 4(A B) =log (ab) log (a+b) + L cot IC, and the side c is then obtained from the formula log c=log a+L sin CL sin A. (3) The two sides a, b and the angle
L sin B=L sin A+log blog a; this gives two supplementary values of the angle B, if b sin A< a. I f b sin .4 > a there is no solution, and if b sin A = a there is one solution. In the case b sin A < a, both values of B give solutions provided b > a, but the acute value only of B is admissible if b < a. The other side c can be then determined as in case (2). (4) If two angles A, B and a side a are given, the angle C is determined from the formula C=7AB and the side b from the formula log b= log +-L sin BL sin A. The area of a triangle is half the product of A Tray off s a side into the perpendicular from the opposite and Quadri- 2b ngle on A, that { s(s S ) (s thus b) (s wecobtain the for the exparessions of na laterals. triangle. A large collection of formulae for the area of a triangle are given in the Annals of Mathematics for 1885 by M. Baker. Let a, b, c, d denote the lengths of the sides AB, BC, CD, DA respectively of any plane quadrilateral and A+C2a; we may obtain an expression for the area S of the quadrilateral in terms of the sides and the angle a. We have 2S=ad sin A+bc sin(2aA) and I(a2+(12b2c2)=ad cos Abe cos (2aA); hence 4S'+1(a2+d2b2c'-)2=a2d"+b2c"2abcd cos 2a. If 2s = a + b+ c + d, the value of S may be written in the form S= {s(sa)(sb)(sc)(sd) abcd costa}i. Let R denote the radius of the circumscribed circle, r of the in- scribed, and ri, r2, ra of the escribed circles of a triangle Radii of Ch- ABC; the values of these radii are given by the followcumscribed, ing formulae: Inscribed R=abc/4S=a/2 sin A, andEscribed r=S/s=(sa)tan IA =4R sin IA sin zB sin IC, Circles of a r1=S/(sa)=s tan A=4R sin IA cos 2B cos IC. Triangle. (1) (3) cos c=cos a cos b sin b=sin B sin c sin a=sin A sin c tan a =tan A sin b= tan c cos B tan b =tan B sin a =tan c cos A cos c=cot A cot B cos A =cos A sin B cos B=cos b sin A (4) A E FIG. 6. B 276 By division we have A 5 sin l(+eb) sin l(a+bc) (7) tan 2 = sin l(b+ca) sin l(a+b+c) S and by multiplication sinA=2{sin (a+b+c) sin (b+ca) sin l(c+ab) sin 1(a+bc){l sin b sin c =1 i cos' a cos' b cost c +2 cos a cos b cos c}; sin b sin c. Hence the quantity k in (3) is {icos' acos'bcosec+2 cos a cos b cos c}i/sin a sin b sin c. (8) Of Half- Apply the polar triangle transformation to the formulae sides. (5), (6), (7) (8) and we obtain a 1 cos l(A+CB) cos l(A+BC cos 2 = sin B sin C a cos l(B+CA) cos 1(A+B+C sng = i sin B sin C a c cos l(B+CA) cos 1(A+B+C ; t s tan= cos l(A+CB) cos l(A+BC If k' =1 i cos'A cos'Bcos2C2 cos A cosBcosC} 1/sin A sin B sinC, we have. kk' (12) io. Let E be the middle point of AB; draw ED at right angles to AB to meet AC in D; then DE Delambre 's bisects the angle ADB. Formulae. Let CF bisect the angleEnd of Article: OBC If you wish, you can link directly to this article.
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