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inclination of the two planes GBC, GBA, which (art. 6) is the spherical angle at B; consequently the spherical angle at B is greater than the angle at B made by the chords AB, CB. In like manner, the spherical angles at A and C, are greater than the respective angles made by the chords meeting at those points. Consequently, the sum of the three angles of the spherical triangle ABC, is greater than the sum of the three angles of the rectilinear triangle made by the chords AB, BC, AC, that is, greater than two right angles. Q. E. 1° D.
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 conse quently their sum less than six right angles. a. E. 2oD.
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 máy 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 the least angle. But the brevity that our plan requires, VOL. III.
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.
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.
From B, A, and c, as poles, let the arcs DF, DE, FE, be described, and by their intersections form another spherical triangle DEF; either side, as DE, of this triangle, is the supplement of the measure of the angle A at its pole; and either angle, as D, has for its- measure the supplement of the side AB.
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 IL 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 equal 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 ALBG 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 DEEFDF + A+B+c=3 x 180°=540°. But (th. 2), DE + EF + DF < 360°: consequently A + B + c > 180°. Thus the first part of theorem 3 is very compendiously demonstrated.
Cor. 3. This theorem suggests mutations that are some times 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 Stevin, 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 semi
circles, and if the whole circles be consisidered, 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 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.
In Every Spherical Triangle, the following proportion obtains, viz, As Four Right Angles (or 360°) to the Surface of a 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. Complete one of its sides as BC into the circle BCEF, which may be supposed to bound the upper hemisphere. Prolong also, at 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 AEF, 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, p for that of the triangle BAF, q for that of CAE, and a for that of the proposed triangle ABC. Then á 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 A; a and p together make up the lune included between the semicircles CAF, CBF, making the angle c: a and q together make up the spheric lune included between the semicircles BCE, BAE, making the angle B. And the surface 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
A:: is: a + m.
180° : B ::
is: a + 9.
180°: c :: is: a + p.
Whence, 180°: A+B+C::s: 3a+m+p+q=2a ++s; and consequently, by division of proportion,
as 180°: A + B + C - 180° :: is: 2a + 4s - s = 2a;
or, 180°: A + B + C-180° :: 48: a = s. Q. E. n*.
Cor. 1. Hence the excess of the three angles of any spherical triangle above two right angles, termed technically the spherical excess, furnishes a correct measure of the surface of that triangle.
Cor. 2. If r3.141593, and d the diameter of the sphere, then is æd2. = the area of the spherical
*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.
Cor. 3. Since the length of the radius, in any circle, is equal to the length of 57-2957795 degrees, measured on the circumference of that circle; if the spherical excess be multiplied by 57-2957795, the product will express the surface of the triangle in square degrees.
Cor. 4. When a 0, then A + B +c=180': and when a = 4s, then A + B + C = 540°. Consequently the sum of the three angles of a spherical triangle, is always between 2 and 6 right angles: which is another confirmation of th. 3.
Cor. 5. When two of the angles of a spherical triangle are right angles, the surface of the triangle varies with its third angle. And when a spherical triangle has three right angles its surface is one-eighth of the surface of the sphere.
Remark. Some of the uses of the spherical excess, in the more extensive geodesic operations, will be shown in the following chapter. The mode of finding it, and thence the area when the three angles of a spherical triangle are given, is obvious enough; but it is often requisite to ascertain it by means of other data, as, when two sides and the included angle are given, or when all the three sides are given. In the former case, let a and b be the two sides, c the included angle, and E the spherical excess: then is cot E, When the three sides a, b, c, are given, the spherical excess may be found by the following very elegant theorem, discovered by Simon Lhuillier :
cot a cot b + cos c sin c
tan = √(tan ++. tan +.tan -. tan=a+b+c).
4 The investigation of these theorem would occupy more space than can be allotted to them in the present volume.
In every Spherical Polygon, or surface included by any num. ber of intersecting great circles, the subjoined proportion obtains, viz, As Four Right Angles, or 360°, to the Surface of a Hemisphere; or, as Two Right Angles, or 180°, to a Great Circle of the Sphere; so is the Excess of the Sum of the Angles above the Product of 180° and Two Less than the Number of Angles of the spherical polygon, to its Area.
For, if the polygon be supposed to be divided into as many triangles as it has sides, by great circles drawn from all the angles through any point within it, forming at that point the vertical angles of all the triangles. Then, by th. 5, it will be