circle, it will measure the shortest distance from any assumed point to the circumference of that circle. 12. Hence again, if upon the sides, AC and BC, (produced if necessary) of a spherical triangle BCA, we take the arcs, CN, CM, each equal 90o, and through the radii GN. GM (figure to art. 7) draw the plane NGM, it is manifest that the point c will be the pole of the circle coinciding with the plane NGM : so that, as the lines GM, GN, are both perpendicular to the common section GC, of the planes AGC, BGc, they measure, by their inclination the angle of these planes; or the arc NM measures that angle, and consequently the spherical angle BCA 13. It is also evident that every arc of a little circle, described from the pole c as centre, and containing the same number of degress as the arc MN, is equally proper for measuring the angle BCA; though it is customary to use only arcs of great circles for this purpose. 14 Lastly, we infer, that if a spherical angle be a right, angle, the arcs of the great circles which form it, will pass mutually through the poles of each other: and that, if the planes of two great circles contain each the axis of the other, or pass through the poles of each other, the angle which they include is a right angle. These obvious truths being premised and comprehended, the student may pass to the consideration of the following theorems. THEOREM I. Any Two Sides of a Spherical Triangle are together Greater than the Third. This proposition is a necessary consequence of the truth, that the shortest distance between any two points, measured on the surface of the sphere, is the arc of a great circle passing through these points. THEOREM II. The Sum of the Three Sides of any Spherical Triangle is Less than 360 degrees. For, let the sides AC, BC, (fig. to art. 7) containing any angle A, be produced till they meet again in D: then will the arcs DAC, DBC, be each 180°, because all great circles cut each other into two equal parts: consequently DAC + DBC= 360°. But (theorem 1) DA and DB are together greater than the third side AB of the triangle DAB; and therefore, since 360°, the sum CA + CB + AB is less CA+CB+DA+ DB than 360°. Q. E. D. THEOREM III. The Sum of the Three Angles of any Spherical Triangle is always Greater than Two Right Angles, but less than Six. 1. The first part of this theorem is demonstrated in cor. 2 of THE. IV. following. 2. The angle of inclination of no two of the planes can be so great as two right angles; because, in that case, the two planes would become but one continued plane, and the arcs, instead of being arcs of distinct circles, would be joint arcs of one and the same circle. Therefore, each of the three spherical angles must be less than two right angles; and consequently their sum less than six right angles. Q E. D. Cor. 1. Hence it follows, that a spherical triangle may have all its angles either right or obtuse; and therefore the knowledge of any two right angles is not sufficient for the determination of the third. Cor. 2. If the three angles of a spherical triangle be right or obtuse, the three sides are likewise each equal to,or greater than 90° and, if each of the angles be acute, each of the sides is also less than 90°; and conversely. Scholium. From the preceding theorem the student may clearly perceive what is the essential difference between plane and spherical triangles, and how absurd it would be to apply the rules of plane trigonometry to the solution of cases in spherical trigonometry. Yet, though the difference between the two kinds of triangles be really so great, still there are various properties which are common to both, and which may be demonstrated exactly in the same manner. Thus, for example, it might be demonstrated here, (as well as with regard to plane triangles in the elements of Geometry, vol. 1) that two spherical triangles are equal to each other, 1st. When the three sides of the one are respectively equal to the three sides of the other. 2dly. When each of them has an equal angle contained between equal sides: and, 3dly. When they have each two equal angles at the extremities of equal bases. It might also be shown, that a spherical triangle is equilateral, isosceles, or scalene, according as it hath three equal, two equal. or three unequal angles: and again, that the greatest side is always opposite to the greatest angle, and the least side to to the least angle. But the brevity that our plan requires, compels us merely to mention these particulars. It may be added, however, that a spherical triangle may be at once right-angled and equilateral; which can never be the case with a plane triangle. THEOREM IV. If from the Angles of a Spherical Triangle, as Poles, there be described, on the Surface of the Sphere, Three Arcs of Great Circles, which by their Intersections form another Spherical Triangle; Each Side of this New Triangle will be the Supplement to the Measure of the Angle which is at its Pole, and the Measure of each of its Angles the Supplement to that Side of the Primitive Triangle to which it is Opposite. Let the sides AB, AC, BC, of the primitive triangle, be produced till they meet those of the triangle DEF, in the points I, L, M, N, G, K : then, since the point A is the pole of the arc DILE, the distance of the points A and E (measured on an arc of a great circle) will be 90°; also, since c is the pole of the arc EF, the points c and E will be 90° distant: consequently (art. 8) the point E is the pole of the arc AC. In like manner it may be shown, that F is the pole of BC, and D that of AB. This being premised, we shall have DL 90°, and IE =90€ whence DL + IE=DL + EL + IL = DE + IL = 180°. Therefore DE = 180°, IL: that is, since 1L is the measure of the angle BAC, the arc DE is the supplement of that measure. Thus also may it be demonstrated that EF is equal the supplement to MN, the measure of the angle BCA, and that DF is eqaul the supplement to GK, the measure of the angle ABC: which constitutes the first part of the proposition. 2dly. The respective measures of the angles of the triangle DEF are supplemental to the opposite sides of the triangles ABC. For, since the arcs AL and BG are each 90°, therefore is AL + BG GL + AB = 180°; whence GL = 180° — AB; that is the measure of the angle D is equal to the supplement to AB. So likewise may it be shown that AC, BC, are equal to the supplements to the measures of the respectively opposite angles E and F. Consequently, the measures of the angles of the triangle DEF are supplemental to the several opposite sides of the triangle ABC. Q. E, D. Cor. 1. Hence these two triangles are called supplemental or polar triangles. Cor. 2 Since the three sides DE, EF, DF, are supplements to the measures of the three angles, A, B, C ; it results that DE + EF + DF + A + B + C = 3×180o 540° But (th. 2), DE + EF + DF < 360°: consequently A + B + C > 180°. Thus the first part of theorem 3 is very compendiously de monstrated. Cor 3 This theorem suggests mutations that are sometimes of use in computation-Thus, if three angles of a spherical triangle are given, to find the sides: the student may subtract each of the angles from 180°, and the three remainders will be the three sides of a new triangle; the angles of this new triangle being found, if their measures be each taken from 180°, the three remainders will be the respective sides of the primitive triangle, whose angles were given. Scholium. The invention of the preceding theorem is due to Philip Langsberg. Vide, Simon Steven, liv. 3, de la Cosmographie, prop. 31 and Alb Girard in loc. It is often however treated very loosely by authors on trigonometry: some of them speaking of sides as the supplements of angles, and scarcely any of them remarking which of the several triangles formed by the intersection of the arcs DE, EF, DF, is the one in question. Besides the triangle DEF, three others may be formed by the intersection of the semicircles, and if the whole circles be considerer', there will be seven other triangles formed. But the proposition only obtains with regard to the central triangle (of each hemisphere), which is distinguished from the three others in this, that the two angles A and F are situated on the d F E D same side of BC, the two в and E on the same side of Ac, and the two C and D on the same side of AB. THEOREM V. In Every Spherical Triangle, the following proportion obtains, yiz, As Four Right Angles (or 360°) to the surface of a Hemisphere; Hemisphere; or, as Two Right Angles (or 180°) to a Great Circle of the Sphere; so is the Excess of the three angles of the Triangle above Two Right Angles, to the Area of the triangle. Let ABC be the spherical triangle. Com- F plete one of its sides as ec into the circle BCEF, which may be supposed to bound E P B the upper hemisphere. Prolong also, at mA a Dm both ends, the two sides AB, AC, until they form semicircles estimated from each angle, that is, until BAE = ABD = CAF = ACD=180°. Then will CBF 180°-BFE; and consequently the triangle AFF, on the anterior hemisphere will be equal to the triangle BCD on the opposite hemisphere. Putting m, m' to represent the surface of these triangles, f for that of the triangle BAF, q for that of CAE, and a for that of the proposed triangle ABC. Then a and m' together (or their equal a and m together) make up the surface of a spheric lune comprehended between the two semicircles ACD, ABD, inclined in the angle aa and ʼn together make up the lune included between the semicircles CAF, CBF, making the angle c: a and 9 together make up the spheric lune included between And the surthe semicircles BCE, BAE, making the angle в. Face of each of these lunes, is to that of the hemisphere, as the angle made by the comprehending semicircles, to two right angles. Therefore, putting is for the surface of the hemisphere, we have : 180°: A :: S a + m. 180°: C : is: a + p. Whence, 180°: A+B+C :: 45: 3a + m + p + q = 2a +}s; and consequently, by division of proportion, as 180°: A+B+C - 180°:: s: 2a + 1s — ‡s = 2a; - or, 180°: A+B+C- 180°: is: a = 15.Q. E. D.* A+B+C-180o 360° Cor.1 Hence the excess of the three angles of any sphererical triangle above two right angles, termed technically the * This determination of the area of a spherical triangle is due to Albert Girard (who died about 1633). But the demonstration now commonly given of the rule was first published by Dr. Wallis. It was considered as a mere speculative truth, until General Roy, in 1787, employed it very judiciously in the great Trigonometrical Survey, to correct the errors of spherical angles. See Phil. Trans. vol. 80, and the next chapter of this volume. spherical |