arcs. Then P will be the pole of the small circle described about ABC. For draw PA, PB, PC; then from the right-angled triangles PCD and PBD it follows that PB=PC; and from the right-angled triangles PCE and PAE it follows that PA = PC; hence PA = PB = PC. Also the angle PAB = the angle PBA, the angle PBC = the angle PCB, and the angle PCA = the angle PAC; therefore PCB + A = 3 (1 + B+C), and PCB=S- A. Let PC = R. Now tan CD= tan CP cos PCD, (Art. 62), thus tan L a= tan R cos (S – A), = tan 1 therefore tan R= ...(1). The value of tan R may be expressed in various forms; thus if we substitute for tan from Art. 49, we obtain 2 a cos S cos S ~{cos مرز tan R= (2). (S – A) cos (S' – B) cos (8 – N Again cos (S – A) = cos {2 (B+C) - 1 A} = cos (B+C) cos1 A + sin } (B+C) sin A sin } A cos } A {cos (6+c) + cos } (6 - c)}, (Art. 54,) cos a sin } a tan R . (3). Substitute in the last expression the value of sin A from Art. 46; thus 2 sin } a sin ? b sin c tan R= W{siu s sin (8 – a) sin (8 – b) sin (8 – c)} 2 sin a sin } 6 sin c b ....... .(4). n It may be shewn, by common trigonometrical formulæ, that 4 sin } a sin 1 b sin fc=sin (8-a) + sin (8 – 6) + sin (8-c) – sin 8 ; ( hence we have from (4) 1 tan R 2n sin (8 – a) + sin (8 – 6) + sin (8 –c) – vin o}....(5) . S 93, To find the angular radii of the small circles described round the triangles associated with a given fundamental triangle. Let R, denote the radius of the circle described round the triangle formed by producing AB and AC to meet again at A'; similarly let R, and R, denote the radii of the circles described round the other two triangles which are similarly formed. Then we may deduce expressions for tan R,, tan R,, and tan R, from those found in Art. 92 for tan R. The sides of the triangle A'BC are a, 7 – b, , and its angles are A, 1-B, 7-C'; hence if 8 =} (a+b+c) and S= } (A + B + C) we shall obtain from Art. 92 tania (1), tan R cos S Similarly we may find expressions for tan R, and tan R. 94. Many examples may be proposed involving properties of the circles inscribed in and described about the associated triangles. We will give one that will be of use hereafter. To prove that 1 Ano We have Ano=1-cos® a - cosa 6 - cos c+ 2 cos a cos 6 cos c; therefore (sin a + sin b + sin c)– 4n* = 2 (1 = 2(1 + sin a sin 6 + sin b sin c + sin c sin a - cos a cos b.cos sc) 2m {si sin 8+ sin (8 – a) + sin (8 – 6) + sin (8–c)}; Also cotr + tan R 1 2n S and by squaring both members of this equation the required result will be obtained. For it may be shewn by reduction that sino s + sino (s – a) + sino (8 – 6) + sino (8 — c) = 2 – 2 cos a cos 6 cos c, and sin s sin (s-a) + sin s sin (8 – 6) + sin s sin (8-c) + sin (8 - a) sin (8-6) + sin (s – 6) sin (8 – c) + sin (8 — c) sin (8 - a) sin a sin b + sin b sin c + sin c sin a. 1 с T 95. In the figure to Art. 89, suppose DP produced through P to a point A' such that DA' is a quadrant, then A' is a pole of BC, and PA' = -r; similarly, suppose EP produced through P 2 to a point B' such that EB' is a quadrant, and FP produced through P to a point " such that FC" is a quadrant. Then A'B'C' is the polar triangle of ABC, and PA' = PBP = PC' = 2 —-. : Thus P is the pole of the small circle described round the polar triangle, and the angular radius of the small circle described round the polar triangle is the complement of the angular radius of the F T T. S. T. small circle inscribed in the primitive triangle. And in like manner the point which is the pole of the small circle inscribed in the polar triangle is also the pole of the small circle described round the primitive triangle, and the angular radii of the two circles are complementary. EXAMPLES. 3 3 1 2 + tanr, + tan , 1 2 In the following examples the notation of the Chapter is retained. Shew that in any triangle the following relations hold contained in Examples 1 to 7: 1. Tanr, tan r, tan rg tan r sino s. 2. Tan R+ cot r= tan R, + cot r, = tan R, + cotr, tan R, + cot rz = 2(cot r + cotr, + cotr, + cotra). 3. Tano R+tan' R, + tan' R, + tano R, cot* r + cotor, + cot’ r, + coté rg. Tan ri 4. – tan no = (1 + cos a + cos b + cos c). cotr, + cot r, + cot ra-cot r 5. Cosecʻr=cot(s-a)cot (8–6)+cot(8–6)cot (8–c)+cot(8–c)(8–a). 6. Cosecʻr, =cot (8 – b) cot(8-c)-cots cot (s—b)-cotscot(8-c). 7. Tan R, tan R, tan R, = tan R sec S. 8. Shew that in an equilateral triangle tan R=2 tan r. 9. If ABC be an equilateral spherical triangle, P the pole of the circle circumscribing it, Q any point on the sphere, shew that cos QA + cos QB + cos QC = 3 cos PA cos PQ. 10. If three small circles be inscribed in a spherical triangle having each of its angles 120°, so that each touches the other two as well as two sides of the triangle, shew that the radius of each of the small circles = 30°, and that the centres of the three small circles coincide with the angular points of the polar triangle. = 3 = VIII. AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 96. To find the area of a Lune. A Lune is that portion of the surface of a sphere which is comprised between two great semicircles. CDEF B Let ACBDA, ADBEA be two lunes having equal angles at A; then one of these lunes may be supposed placed on the other so as to coincide exactly with it; thus lunes having equal angles are equal. Then by a process similar to that used in the first proposition of the Sixth Book of Euclid it may be shewn that lunes are proportional to their angles. Hence since the whole surface of a sphere may be considered as a lune with an angle equal to four right angles, we have for a lune with an angle of which the circular measure is A, Suppose r the radius of the sphere, then the surface is 4top? (Integral Calculus, Chap. vii.); thus |