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OBC

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Originally appearing in Volume V27, Page 276 of the 1911 Encyclopedia Britannica.
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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 of the sphere. By drawing a perpendicular from A on OB, we may in a similar manner show that AM= r sin c sin B, 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 (I) gives 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 through A, B, C to meet the opposite sides in D, E, F respectively, and if these arcs pass through a point, the segments of the sides satisfy the relation sin BD sin CE sin AF=sin CD sin AE sin BF; and conversely if this relation is satisfied the arcs pass through a point. From this theorem it follows that the three perpendiculars from the angles on the opposite sides,' the three bisectors of the angles, and the three arcs from 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 forSine perpendicular to BC, it may be shown that BE and Cosine = 1(a + c — b) and CE = 1(a + b — c), and that of Halt the angles BDE, ADC are supplementary. We have Angles. 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+c—b), and sin CD sin CDE=sin CE=sin 2(a+b—c); therefore sA csin 2(a+c—b) sin 1(a+b—c) z 2 sin b sin c (5) in — Apply this formula to the associated triangle of which ,r—A, ,r—B, C are the angles and ,r—a, ,r—b, c are the sides; we obtain A )sin z(b+c—a) sin 1(a.+b+c) ; (6) - = the formula cos sin b sin c ) (2) The two sides a, b and the included angle C being given, the angles A, B can be determined from the formulae A+B=r—C, L tan 4(A —B) =log (a—b) —log (a+b) + L cot IC, and the side c is then obtained from the formula log c=log a+L sin C—L sin A. (3) The two sides a, b and the angle A being given, the value of sin B may be found by means of the formula L sin B=L sin A+log b—log 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=7—A—B and the side b from the formula log b= log ¢+-L sin B—L 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+C—2a; 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(2a—A) and I(a2+(12—b2—c2)=ad cos A—be cos (2a—A); hence 4S'+1(a2+d2—b2—c'-)2=a2d"+b2c"—2abcd cos 2a. If 2s = a + b+ c + d, the value of S may be written in the form S= {s(s—a)(s—b)(s—c)(s—d) —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=(s—a)tan IA =4R sin IA sin zB sin IC, Circles of a r1=S/(s—a)=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(¢+e—b) sin l(a+b—c) (7) tan 2 = sin l(b+c—a) sin l(a+b+c) S and by multiplication sinA=2{sin (a+b+c) sin (b+c—a) sin l(c+a—b) sin 1(a+b—c){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 {i—cos' a—cos'b—cosec+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+C—B) cos l(A+B—C cos 2 = sin B sin C a —cos l(B+C—A) cos 1(A+B+C sng = i sin B sin C a c —cos l(B+C—A) cos 1(A+B+C ; t s tan= cos l(A+C—B) cos l(A+B—C If k' =1 i —cos'A —cos'B—cos2C—2 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 angle
End of Article: OBC
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