5. The right-angled spherical triangle Gae has three right angles, for A is a right angle by the proposition, and A is the pole of the circle Eg, therefore aeg and age are right angles. (H. 135.) Q. E. D. (P) COROLLARY I. In any right-angled triangle, if the two legs be of the same species, viz. both acute, or both obtuse, the hypothenuse will be less than a quadrant, and the contrary. For bc the hypothenuse, or side opposite to the right angles A and d, is common to the triangle bac having two acute angles, and to the triangle bpc having two obtuse angles. (Q) CorollaRY II. In any right-angled spherical triangle, if the legs be of different species, the hypolhenuse will be greater than a quadrant, and the contrary: but one of the legs, at least, will be a quadrant, if the hypothenuse be a quadrant. For, in the triangle icd, right-angled at c, ci is less than a quadrant and cd greater, from what has already been demonstrated; but since d is the pole of the circle EHG, Dh is a quadrant, therefore di the hypothenuse is greater than a quadrant. And when cı becomes equal to ce, di will become equal to DE; but DE is a quadrant, therefore if the hypothenuse be a quadrant, one of the legs, at least, will be a quadrant. But both the legs will be quadrants when all the angles are right angles, as in the triangle AEG. (R) COROLLARY III. The sides containing the right angle of any triangle, are always of the same species as their opposite angles, and the angles are of the same species as their opposite sides. For in the triangle cas where the two angles are acute, the opposite sides AB and ac are acute; and in the triangle cdB where the two angles are obtuse, the opposite sides BD and cp are obtuse. (S) COROLLARY IV. If the hypothenuse be less than 90°, the kegs are of the same species as their adjacent angles. If the hypothenuse be greater than 90°, the legs and their adjacent angles are of different species. This is evident from the triangles BAC, CDB, and dci. (T) COROLLARY V. According as the hypothenuse, and one side (or its opposite angle) are of the same, or different species ; the other side (and its opposite angle) will be less or greater than a quadrant. This is also evident, from the triangle dci. PROPOSITION XIV. (U) If any number of arcs of circles CPB, CME, CNF, CDA, CSH, Cog, be drawn from one and the same point c, on the surface of the sphere, to the circumference of a great circle AHGFEB ; the greatest arc will be that which passes through the pole 1. of the great circle AHGFEB; and the least arc will be the supplement adc of the greatest CPB. And that arc which is the nearest to the greatest will be greater than any one more remote ; likewise, that arc which is nearest to the least arc, will be less than any one more remote from it. DEMONSTRATION. Let z be the centre of the sphere, or of the great circle AHGFEB, and as its diameter; from c, the point where all the circles intersect each other, draw ci perpendicular to AB, and join IE, IF, IG and in. Then of all the straight lines which can be drawn from the point i to the circumference of the circle AHGFEB, that line 1B, is the greatest which passes through the centre z, and its opposite line ia is the least; that line ie, which is nearest to IB, is greater than the line if which is more remote; and the line in which is nearest to IA, is shorter than the line ig which is more remote. (Euclid 7 of III.) Let AC, HC, GC, FC, &c. be joined by straight lines; then will CIB, CIE, CIF,crg, cih and D cia be all right-angled plane triangles, having the perpendicular ci common to each; VZ therefore that triangle CIB, А. which has the greatest base IB, will have the greatest hy H E pothenuse Bc; and that tri F angle cia which has the least base ia, will have the least hypothenuse Ac, &c.; of the others, EC will be greater than pc, and HC less than GC. But these several hypothenuses are all chords of equal circles, and the greater chord cuts off the greater arc, or circumference, therefore CPB is the greatest arc, and adc the least; CME is greater than any, and csh is less than cog. Q. E. D. P PROPOSITION XV. (W) If the angles at the base of an oblique-angled spherical triangle be of the same species, viz. both acute, or both obtuse, a perpendicular from the vertical angle upon the base will fall within the triangle; if they be of different species, it will fall without. DEMONSTRATION. In the rightangled triangle CDB, cd is of the same species as CBD (R. 145.), and in the right-angled triangle CDA, CD is of the same species as CAD, II. therefore the perpendicular in the first figure falls between the angles A and B. I. In the second figure, CBD and CBA are supplements of each other (M. 194.), and the perpendicular cd falls opposite to both CBD and cad being of the same species, therefore it cannot fall opposite to CBA at the same time, hence it must necessarily fall without the triangle. Q. E. D. (X) COROLLARY. In an isosceles triangle, the equal angles are of the same species as their adjacent, or opposite sides. For then ad=DB (K. 143.) and these segments are always acute, since their sum can never be equal to a semi-circle. (C. 133.) But according as DB and Dc, or AD and Dc, are of the same, or different species, Bc or its equal ac, is acute or obtuse; or since the segments are always acute, bc and ac will be of the same species as DC, but dc is of the same species as A and B, therefore the sides are of the same species as their adjacent or opposite base angles. PROPOSITION XVI. (Y) If the LEAST two sides of a spherical triangle be of the same species, the perpendicular drawn from their included angle upon the third side, will fall within the triangle. Or, if the third side be less than either of the others, they being of the same species, but greater than their supplements, the perpendicular will fall within the triangle, and the less segment of the base and less vertical angle will be adjacent to the greater side, and the greater segment, and greater vertical angle will join the less side, provided the sum of the sides be greater than 180°. DEMONSTRATION. Let ABC be the triangle, cd a perpendicular upon its E base AB; make BN equal to the hypothenuse BC, and an equal to the hypothenuse AC, then BCN and AcI are isos 12 celes triangles ; the angle bnc is of the same species as BC, and the angle AIC is of the same species as Ac. (X.147.). But Ac and Bc are of the same species, by the hypothesis, therefore the angles BNC and Aic are of the same species; consequently cd falls between them (W. 146.) and of course within the triangle ABC. SECONDLY. In the triangle AVB, where av and bv are each greater than AB, the least two sides of the supplemental triangle FVE, which by the hypothesis are each less than AB or FE, are of the same species, and consequently vo falls between them, therefore vd falls between Av and Bv. But av is greater than BV, and ad is less than db; consequently avd is less than BVD. Q. E. D. (Z) COROLLARY I. The perpendicular let fall on the base of an oblique spherical triangle is either less or greater than each side. For since cd is perpendicular to ab, it must pass through p the pole of AB (I. 134.); and if c be below the pole P, that is if cd be less than a quadrant, it will be the least of all the lines that can be drawn from the point c to the circumference of the circle ABEF, and cg will be the greatest. (U. 145.) If c the vertex of the triangle Abc fall above the pole p, suppose in v, then vd will be the perpendicular, and of all the arcs that can be drawn from v upon the arc AB, vp is the greatest (U. 145.), therefore the perpendicular is either less or greater than each side ; when it becomes equal to one of the sides, the triangle will be right angled, as BDC. If the perpendicular fall without the triangle, as in Boc, or BOV, the demonstration will hold equally true. (A) COROLLARY II. When the perpendicular falls without the triangle, it may be either less or greater than a quadrant ; or less or greater than each side. In the triangle Boc, either CD or cu may be esteemed the perpendicular upon the base os continued; the former cd is less than a quadrant, and less than either co or cs, and the latter cg is greater. (B) COROLLARY III. When the perpendicular falls without the triangle, the less perpendicular is next the less side, and the greater perpendicular is next the greater side; and either of them may be considered as the proper perpendicular or the base produced. In the triangle Boc, cd the less perpendicular is next the less side co, and co the greater perpendicular is next the greater side BC, and either of them may be used for determining the several parts of the triangle Boc, but the one is sometimes more convenient than the other. (C) COROLLARY IV. In any triangle ABC Or BOC, the perpendicular cd falling within the triangle ABC upon the base AB, Or NEAREST perpendicular cd falling without the triangle Boc, is always of the same species as half the sum of the two sides of the triangle ; and if half the sum of the two sides of the triangle be acute, this perpendicular falls nearest the less side, if obtuse, it falls nearest the greater side. In the triangle ABC the perpendicular pc is less than a quadrant, and each side ac and Bc is less than a quadrant, therefore half their sum is less than a quadrant, and the perpendicular cd falls nearest to AC. In the triangle Boc, cd is less than a quadrant, and each of the sides co and cB is less than a quadrant, therefore half their sum is less than a quadrant, and the perpendicular falls nearest And, in the triangles abv and Bov, where the perpendicular vo is greater than a quadrant, each of the sides AV, BV, and ov is greater than a quadrant; consequently half their sun must be greater than a quadrant, and the perpendicular və falls nearest to Av in the former triangle, and to ov in the latter. (D) COROLLARY V. If an oblique spherical triangle have two acute sides and one obtuse, a perpendicular drawn on the longest side will always fall within the triangle (Y. 147.)-But if it has two obtuse sides and one acute, a perpendicular drawn to fall on the longest side may sometimes fall within, and sometimes without the triangle. Take any right-angled triangle bbc, and produce its sides to a semi-circle; now because DB and pc are each less than a quadrant, the hypothenuse Bc will be less than a quadrant (Q. 145.), consequently its supplement cF will be greater than a quadrant. The angles B and c being acute or each less than a right angle, bc is greater than DB (L. 143.), therefore DF is greater than cf. Make Fa equal to Fc, then a great circle as uc may be drawn between A and D; now in the triangle fuc, fu is greater than FC; FC, hc, are of different species, and the perpendicular CD falls without the triangle fuc. In the triangle foc, of is greater than Fc; fc and oc are of different species, and the perpendicular cd falls within the triangle Foc. Therefore there can be no general rule for drawing a perpendicular when the three sides of a triangle are given, especially if two of those sides be obtuse. (E) COROLLARY VI. If the two obtuse sides of an obliqueangled triangle be equal, a perpendicular drawn from their .included angle will fall within the triangle. For then the triangle is isosceles, and the angles at the base are equal to each other. (F. 142.) PROPOSITION XVII. (F) In any spherical triangle, 1. If the three angles be each acute, cach side will be less than a quadrant. 2. If the three angles be each right F angles, each side will be a quadrant. 3. If the three angles be each obtuse, each side will be greater than a quadrant. B DEMONSTRATION. First, Let all the D t A |