103. We may obtain many other formulæ involving trigonometrical functions of the spherical excess. Thus, for example, cos E = cos {} (A+B) − } (π − C')} = cos (A + B) sin 1⁄2 C + sin † (4 + B) cos 1⁄2 C ={c cos § (a + b) sin3 } C′+cos } (a−b) coso 1 C'} sec } c, (Art. 54), b + sin } a sin } 6 (cos" } C′ – sin2 } C)} sec } c = {cosa cos b + sina sin b cos C} secc. Again, it was shewn in Art. 101, that sin E = sin C sina sin 16 sec1c; cos E = = { cosa cos 6 + sin a sin 6 cos C} sec ↓ c (1 + cos a) (1 + cos b) + sin a sin b cos C 4 cosa cosb cos c 1 + cos a + cos b + cos c cos2 a + cos2 1b + cos3 c = 4 cosa cosb cos c 2 cosa cosb cos c In (3) put 1-2 sin E for cos E; thus sin'E= 1+2 cosa cos bcos c cos2 a- cos3 b-cos3 c 4 cosa cosb cos c By ordinary development we can shew that the numerator of the above fraction is equal to 4 sin 8 sin † (8 − a) sin † (8 — b) sin † (s – c) ; therefore sin❜E= sins sin (sa) sin (s—b) sin (sc) Similarly cos E= cos 1⁄2s cos (s - a) cos † (s – b) cos (s – c) .(5). From this result we can deduce two other results, in the same manner as (4) and (5) were deduced from (3); or we may observe that the right-hand member of (6) can be obtained from the right-hand member of (3) by writing - a and b for a and b respectively, and thus we may deduce the results more easily. We shall have then sin' (C-E) Coss sin (8 − a) sin (s – b) cos § (8 — c) sin = sina sin cos c 1. Find the angles and sides of an equilateral triangle whose area is one-fourth of that of the sphere on which it is described. 2. Find the surface of an equilateral and equiangular spherical polygon of n sides, and determine the value of each of the angles when the surface equals half the surface of the sphere. 4. If the angle C of a spherical triangle be a right angle, shew that sin E = sina sinb secc, cos E = cosa cos b secc. 5. If the angle C be a right angle, shew that 7. The sum of the angles in a right-angled triangle is less than four right angles. 8. Draw through a given point in the side of a spherical triangle an arc of a great circle cutting off a given part of the triangle. If the angles of a spherical triangle be together equal to four right angles 10. If r cos2 a + cos2 b + cos2 1 c = = 1. r be the radii of three small circles of a sphere of radius r which touch one another at P, Q, R, and A, B, C be the angles of the spherical triangle formed by joining their centres, sin s= sin † E sin (4 – † E) sin (B – ‡ E) sin (C' – † E)}1 2 sin A sin B sin C 12. Given two sides of a spherical triangle, determine when the area is a maximum. 13. Find the area of a regular polygon of a given number of sides formed by arcs of great circles on the surface of a sphere; and hence deduce that, if a be the angular radius of a small circle, its area is to that of the whole surface of the sphere as versin a to 2. 14. A, B, C are the angular points of a spherical triangle; A', B', C' are the middle points of the respectively opposite sides. If E be the spherical excess of the triangle, shew that 15. If one of the arcs of great circles which join the middle points of the sides of a spherical triangle be a quadrant, shew that the other two are also quadrants. IX. ON CERTAIN APPROXIMATE FORMULÆ. 104. We shall now investigate certain approximate formula which are often useful in calculating spherical triangles when the radius of the sphere is large compared with the lengths of the sides of the triangles. 105. Given two sides and the included angle of a spherical triangle, to find the angle between the chords of these sides. E C B Let AB, AC be the two sides of the triangle ABC; let O be the centre of the sphere. Describe a sphere round A as a centre, and suppose it to meet 10, AB, AC at D, E, F respectively. Then the angle EDF is the inclination of the planes OAB, OAC, and is therefore equal to 4. From the spherical triangle DEF cos EF = cos DE cos DF + sin DE sin DF cos A ; DE = (π-c), and therefore DF=1⁄2 (π—b); cos EF = sin 6 sin c + cos b cosc cos A. If the sides of the triangle are small compared with the radius of the sphere, EF will not differ much from A; suppose EF-A-0, then approximately cos EF = cos A + 0 sin A ; |