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This truth is evident from Prop. IX, where it was shown, that with three given sides AB, AC, BC, there can only be two triangles ACB, ABD, differing as to the position of their parts, and equal as to the magnitude of those parts. Hence those two triangles, having all their sides re- D4 spectively equal in both, must either be absolutely equal, or at least symmetrically so; in either of which cases, their corres


ponding angles must be equal, and lie opposite to equal sides.


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In every isosceles spherical triangle, the angles opposite the equal sides are equal; and conversely, if two angles of a spherical triangle are equal, the triangle is isosceles.

First. Suppose the side AB-AC; we shall have the angle C=B. For, if the arc AD be drawn from the vertex A to the middle point D of the base, the two triangles ABD, ACD, will have all the sides of the one respectively equal to the corresponding sides of the other, namely, AD common, BDDC, and AB= AC: hence by the last Proposition, their an- B gles will be equal; therefore, B=C.


Secondly. Suppose the angle B-C; we shall have the side AC AB. For, if not, let AB be the greater of the two; take BO-AC, and draw OC. The two sides BO, BC, are equal to the two AC, BC; the angle OBC, contained by the first two is equal to ACB contained by the second two. Hence the two triangles BOC, ACB, have all their other parts equal (Prop. X.); hence the angle OCB-ABC: but by hypothesis, the angle ABC-ACB; hence we have OCB ACB, which is absurd; hence it is absurd to suppose AB different from AC; hence the sides AB, AC, opposite to the equal angles B and C, are equal.

Scholium. The same demonstration proves the angle BAD= DAC, and the angle BDA ADC. Hence the two last are right angles; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to that base, and bisects the vertical angle.


In any spherical triangle, the greater side is opposite the greater angle; and conversely, the greater angle is opposite the greater side.

Let the angle A be greater than the angle B, then will BC be greater than AC; and conversely, if BC is greater than AC, then will the angle A be B greater than B.


First. Suppose the angle A>B; make the angle BAD=B; then we shall have AD=DB (Prop. XIII.): but AD+DC is greater than AC; hence, putting DB in place of AD, we shall have DB+DC, or BC>AC.

Secondly. If we suppose BC>AC, the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC=AC; if BAC were less than ABC, we should then, as has just been shown, find BC AC. Both these conclusions are false: hence the angle BAC is greater than ABC.


If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they will also be mutually equilateral.

Let A and B be the two given triangles; P and Q their polar triangles. Since the angles are equal in the triangles A and B, the sides will be equal in their polar triangles P and Q (Prop. VIII.): but since the triangles P and Q are mutually evuilateral, they must also be mutually equiangular (Prop. XII.); and lastly, the angles being equal in the triangles P and Q, it follows that the sides are equal in their polar triangles A and B. Hence the mutually equiangular triangles A and B are at the same time mutually equilateral.

Scholium. This proposition is not applicable to rectilineal triangles; in which equality among the angles indicates only proportionality among the sides. Nor is it difficult to account for the difference observable, in this respect, between spherical and rectilineal triangles. In the Proposition now before us,


as well as in the preceding ones, which treat of the comparison of triangles, it is expressly required that the arcs be traced on the same sphere, or on equal spheres. Now similar arcs are to each other as their radii; hence, on equal spheres, two triangles cannot be similar without being equal. Therefore it is not strange that equality among the angles should produce equality among the sides.

The case would be different, if the triangles were drawn upon unequal spheres; there, the angles being equal, the triangles would be similar, and the homologous sides would be to each other as the radii of their spheres.


The sum of all the angles in any spherical triangle is less than six right angles, and greater than two.

For, in the first place, every angle of a spherical triangle is less than two right angles: hence the sum of all the three is less than six right angles.

Secondly, the measure of each angle of a spherical triangle is equal to the semicircumference minus the corresponding side of the polar triangle (Prop. VIII.); hence the sum of all the three, is measured by the three semicircumferences minus the sum of all the sides of the polar triangle. Now this latter sum is less than a circumference (Prop. III.); therefore, taking it away from three semicircumferences, the remainder will be greater than one semicircumference, which is the measure of two right angles; hence, in the second place, the sum of all the angles of a spherical triangle is greater than two right angles.

Cor. 1. The sum of all the angles of a spherical triangle is not constant, like that of all the angles of a rectilineal triangle; it varies between two right angles and six, without ever arriving at either of these limits. Two given angles therefore do not serve to determine the third.

Cor. 2. A spherical triangle may have two, or even three of its angles right angles; also two, or even three of its angles obtuse.


Cor. 3. If the triangle ABC is bi-rectangular, in other words, has two right angles B and C, the vertex A will be the pole of the base BC; and the sides AB, AC, will be quadrants (Prop. V. Cor. 3.)


If the angle A is also a right angle, the triangle ABC will be tri-rectangular; its angles will all be right angles, and its sides quadrants. Two of the tri-rectangular triangles make half a hemisphere, four make a hemisphere, and the tri-rectangular triangle is obviously contained eight times in the surface of a sphere.

Scholium. In all the preceding observations, we have supposed, in conformity with (Def. 1.) that spherical triangles have always each of their sides less than a semicircumference; from which it follows that any one of their angles is always less than two right angles. For, if the side AB is less than a semicircumference, and AC is so likewise, both those arcs will require to be produced, before they can meet in D. Now the two angles ABC, CBD, taken together, are equal to two right angles; hence the angle ABC itself, is less than two right angles.


We may observe, however, that some spherical triangles do exist, in which certain of the sides are greater than a semicircumference, and certain of the angles greater than two right angles. Thus, if the side AC is produced so as to form a whole circumference ACE, the part which remains, after subtracting the triangle ABC from the hemisphere, is a new triangle also designated by ABC, and having AB, BC, AEDC for its sides. Here, it is plain, the side AEDC is greater than the semicircumference AED; and at the same time, the angle B opposite to it exceeds two right angles, by the quantity CBD.

The triangles whose sides and angles are so large, have been excluded by the Definition; but the only reason was, that the solution of them, or the determination of their parts, is always reducible to the solution of such triangles as are comprehended by the Definition. Indeed, it is evident enough, that if the sides and angles of the triangle ABC are known, it will be easy to discover the angles and sides of the triangle which bears the same name, and is the difference between a hemisphere and the former triangle.



The surface of a lune is to the surface of the sphere, as the angle of this lune, is to four right angles, or as the arc which measures that angle, is to the circumference.


Let AMBN be a lune; then will surface be to the surface of the sphere as the angle NCM to four right angles, or as the arc NM to the circumference of a great circle.



Suppose, in the first place, the arc MN to be to the circumference MNPQ as some one rational number is to another, as 5 to 48, for example. The circumference MNPQ being divided into 48 equal parts, MN will contain 5 of them; and if the pole A were joined with the several points of division, by as many quadrants, we should in the hemisphere AMNPQ have 48 triangles, all equal, because all their parts are equal. Hence the whole sphere must contain 96 of those partial triangles, the lune AMBNA will contain 10 of them; hence the lune is to the sphere as 10 is to 96, or as 5 to 48, in other words, as the arc MN is to the circumference.

If the arc MN is not commensurable with the circumference, we may still show, by a mode of reasoning frequently exemplified already, that in that case also, the lune is to the sphere as MN is to the circumference.

Cor. 1. Two lunes are to each other as their respective angles.

Cor. 2. It was shown above, that the whole surface of the sphere is equal to eight tri-rectangular triangles (Prop. XVI. Cor. 3.); hence, if the area of one such triangle is represented by T, the surface of the whole sphere will be expressed by 8T. This granted, if the right angle be assumed equal to 1, the surface of the lune whose angle is A, will be expressed by 2AXT: for,

4: A

8T: 2A×T

in which expression, A represents such a part of unity, as the angle of the lune is of one right angle.

Scholium. The spherical ungula, bounded by the planes AMB, ANB, is to the whole solid sphere, as the angle A is to

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