PROPOSITION III. be described in cross direction to mutually intersect, a new triangle 1. The two triangles shall be mutually polar to each other; and 2: The sides of each triangle shall be supplementary to the angles of the other. 1. With poles A, B, C describe great circles in cross direction with respect to the sides AB, BC, CA of the triangle ABC, mutually intersecting in a, b, c; then ABC, abc will be mutually polar to each other. Since A is the pole of be, the distances Ab, Ac are each a quadrant; and similarly Ba, Bc, Ca, Cb, are also each of them a quadra!it. Ilence, arranging the order of them differently, aB and aC are each of them a quadrant; and therefore a is the pole of BC. Similarly b is the pole of AC, and c that of BA. 2. Since B and C are the poles of ac, ab, taken in cross direction, BC is the measure of the supplement of the angle bac, formed by ab, ac (Prop. II. Schol.), or the side BC is supplementary to the angle bac. In the same manner CA and AB are supplementary to the angles abc and acb. Again, since the triangle ABC is polar to abc, the same reasoning will show that bc is supplementary to BAC, ca to ABC, and ab to ACB. SCHOLIUM. This theorem was discovered by Philip Lansberg, and has since his time been extensively employed in trigonometrical research. It is to be recollected, in the enunciation of this theorem, that by a spherical angle is meant its measure, viz., the circular arc intercepted by its sides upon the circle to which they are both secondaries. It must also be carefully remembered that this supplementary property only holds in the two triangles when they are taken in cross directions. PROPOSITION IV. third, and the difference of any two is less than the third, For, let O be the centre of the sphere upon which the spherical triangle ABC is described, and join OA, OB, OC. Then AOB, BOC, COA, are the plane angles which contain the solid angle at 0. Then of these angles the sum of any two is greater than the third, and the difference of any two is less than the third, (Pls. and Sols., Chap. iv., Prop. 1.). Also, these angles are measured by the arcs of the circles which form the spherical triangle ABC; and hence the propocition is true of these arcs, that is, of the sides of the triangle. O 3 PROPOSITION V. In a spherical triangle, the three sides are together less than an entire circumference, and each side is less than a semicircle, (See Fig. Prop. iv.). 1. For the three angles at O are together less than four right angles (Pls. and Sols., Chap. iv., Prop. 111.), and hence their measures AB, BC, CA are together less than four quadrants: that is, than an entire circumference. 2. Since any two sides AB, BC, are together greater than the third AC, it is obvious that one side, AC, must be less than half the sum of all the three sides AB, BC, CA ; that is, by the preceding case, less than a semicircle PROPOSITION VI. In a spherical triangle, if two of the angles be equal, the sides oppo site to them are equal, and conversely, if two sides be equal, the angles opposite to them are equal. 1. Let the angles CAB, CBA, be equal; then shall BC be equal to CA. For, draw CD perpendicular to the plane AOB and in the same plane draw DE, DF, perpendicular to OA, OB, and join EC, FC. Then (Pls. and Sols., Chap. II., Prop. xu.) CE and CF are perpendicular to OA and OB; and hence the angles CED, CFD are equal to the spherical angles at A and B. But (by hypothesis) these are equal; and hence in the right-angled plane triangles CDE, CDF we have the angles CED, CFD, equal, and the side CD common; wherefore also EC is equal to CF. Again, in the right-angled triangles CEO, CFO, we have EC equal to CF and CO common; hence the angles COE COF, are also equal, and their measures AC, CB, also equal. 2. Let the sides AC, CB be equal, then the angles at A and B will also be equal. For, make the same construction; then, since the sides AC, CB are equal, the angles AOC, COB are equal ; and since OC is common to the two right-angled triangles CEO, CFO, we have EC equal to FC. Wherefore, in the right-angled triangles CED, CFD, we have EC, CD, equal to FC, CD each to each ; and therefore the angles CED, CFD also equal : and these are equal to the spherical angles at A and B. Cor. In every spherical triangle, the greater side and greater angle are opposite. For, let AC be greater than ÇB; then the angle COA is greater than. COB, and therefore EC greater than CF: wherefore pursuing a train of reasoning analogous to that in the proposition, we have CD common to the triangles CDE, CDF, but EC greater than CF, and hence the angle CFD greater than CED, that is, B is greater than A. PROPOSITION VII. all three together less than six right angles. 1. For every dihedral angle is less than two right angles, and hence every spherical angle which has the same measure is less than two right angles. 2. Hence, the three together are less than three times two right angles; that is, less than six right angles. Cor. Hence, contrary to what takes place in plane triangles, there may be two, or even three, of the angles of a spherical triangle either right angles or obtuse angles. PROPOSITION VIII. two right angles. For, let ABC be any spherical triangle, and abc its cross-polar : and draw the arcs Ab, Ac, Ba, Bc, Ca, Cb. Then since A is the pole of bc, the angle bAc is measured by bc; and since BAC and bc are supplementary, the angles BAC and bAc are together equal to two right angles. In like manner ABC and aBc are together equal to two right angles, and likewise BCA and 6Ca are together equal to two rights angles. Hence the six angles ABC, BCA, CAB, aBc, Ca, and cAb are together equal to six right angles. But the angles bAc, cBa, aCó are measured by the sides of the triangle abc, and these are together less than an entire circumference, and the angles themselves, therefore, less than four right angles. Wherefore, the remaining three angles ABC, BCA, and ČAB of the spherical triangle ABC are greater than two right angles. PROPOSITION IX. other, each to each, the angles will be equal which are opposite to the 1. Suppose the sides to follow in the same order, and let the triangles be abc, A'B'C'. Then, as in a plane, the sides ab, bc, ac, will be capable of coincidence respectively with A'B', B'C', C'A', and the conclusion will follow. 2. Suppose them not in the same order, as abc, ABC. Then produce the edges OA, OB, OC, to A', B', C', and let the planes of each two cut the spherical surface in B'C', C'A', A'B'. Then these are equal to BC, CA, AB, respectively, and hence also to bc, ca, ab respectively. But the dihedral angles at OA', OB', OC', are equal to those at OA, OB, OC, since they are vertical to them ; and the dihedral angles at oa, ob, oc, are also equal to those at OA', OB', OC', since they are contained by equal plane angles, or those measured by equal arcs. Whence the dihedral angles, that is, the spherical angles of the triangles ABC, abc are equal, each to each. In like manner, by a mere contra-position of the terms of the argument, the converse case may be proved. SCHOLIUM. By methods correspondent to these, the equalities analogous to those where different combinations of the parts of plane triangles are concerned, may be established. PROPOSITION X. Let two great semicircles cut at right angles, then from any point in one of them innumerable pairs of circles may be drawn, so that each pair shall make equal angles with the other semicircle. Let RPR', RQR' be two semicircles intersecting at right angles ; then through any point P in the former, innumerable circles may be drawn in pairs, as PA, PB, which shall make, with the latter, angles PAR, PBR equal to one another. Bisect the semicircle RQR' in Q; take any two equal arcs AQ, BQ, in RQR' and draw the circles AP, PB. These will make the angles RBP, RAP equal to one another. For, produce the circles AP, BP, and draw QP to meet the completion of the circle RQR', in A', B', Q. Then we have BR + RA = (RQ – QB) + (RQ + QA) = 2RQ = RQR' = * ARA' AR + RA'. Wherefore, A'R = RB, and consequently also, RBP = RA'P = RAP. Whence, taking AQ, QB, of any magnitude, but equal to each other, the angles formed by PA, PB with RQR' will be equal; which establishes the proposition. Cor. 1. If the point P lie (as in the first figure) between R and the pole O of the circle RQQ' the angles PAR, PBR will be acute, and have as their greatest limit the right angles PRQ', PR'Q : but if it be beyond O estimated from R (as in the second figure), the angles PAR, PBR will be obtuse, and have for their least limit the same right angles PRQ', PR'Q. Cor. 2. On the hypothesis of the last corollary, PQR will be the least in the former figure, and the greatest in the latter, of all the angles that can be made by circles passing through P and cutting the circle RQR'. For, draw OEF perpendicular to PA, the one of the circles that passes between PR'and PQ meeting it in E, and RQR' in F. Then, since the angle R'PA is acute, it is less than OEP (Fig. 1), and therefore also OE is less than OP; and as QR OF = { 1, we have EF greater than PR. But AA' is bisected in E, or AE is a quadrant, and at right angles to EF; wherefore A is the pole of EF, and the angle EAF or PĂR is measured by EF. Also PQR is measured by PR; wherefore PQR is less than PAR. In the second figure we may reason similarly ; but it will be more simple to deduce the conclusion from the supplements of the angles. Thus, since R'QP (Fig. 2) is less than R'AP, it follows that PQR is greater than PAR. Cor. 3. The circles PA, PB make equal angles with the circle OP or RR', as is at once obvious. Cor. 4. If from any point P on the surface of a hemisphere bounderl by the circle R'QRQ' circles PR, PB, PQ, PA, PR', etc., be drawn to meet that circle; then the greatest is that which passes through the pole, and the least is that constituting the remainder of the semicircle ; those which pass, one on each side of the greatest, at equal distances from the pole (or which make equal angles on opposite sides with the greatest), are equal, etc., to correspond with Euc. in. 7, 8. PROPOSITION XI. Two sides of a trigngle and the angle opposite to one of them being given, it is required to investigate the number of solutions of which a triangle constructed with these data admits. Let A be the given angle, BC the given side opposite to A, and AC the other given side: it is required to find when the point B can have two positions, and when only one, compatible with the given magnitudes A, a, b. [In Fig. 1, A is acute, and in Fig. 2, A is obtuse.] Produce AB, AC to meet in A', and from C draw CD perpendicular to ABA'. With pole C, and spherical radius equal to any given value of the side a, describe circles to cut ADA', which each will do, as for instance, in two points b, b'. D |