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.

PROPOSITION XVI. THEOREM.

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 sphe rical 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

B

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.

A

C

B

D

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.

E

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.

PROPOSITION XVII. THEOREM.

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 its 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.

M

N

B

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: A8T 2AXT

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

four right angles. For, the lunes being equal, the spherical ungulas will also be equal; hence two spherical ungulas are to each other, as the angles formed by the plancs which bound them.

equal arcs PA, PB, PC (Prop. V.); at the point F, make the angle DFQ-ACP, the arc FQ=CP; and draw DQ, EQ.

The sides DF, FQ, are equal to the sides AC, CP; the angle DFQ=ACP: hence the two triangles DFQ, ACP are equal in all their parts (Prop. X.); hence the side DQ=AP, and the angle DQF APC.

In the proposed triangles DFE, ABC, the angles DFE, ACB, opposite to the equal sides DE, AB, being equal (Prop. XII.). if the angles DFQ, ACP, which are equal by construction, be taken away from them, there will remain the angle QFE, equal to PCB. Also the sides QF, FE, are equal to the sides PC, CB; hence the two triangles FQE, CPB, are equal in all their parts; hence the side QE=PB, and the angle FQE=CPB.

Now, the triangles DFQ, ACP, which have their sides respectively equal, are at the same time isosceles, and capable of coinciding, when applied to each other; for having placed AC on its equal DF, the equal sides will fall on cach other, and thus the two triangles will exactly coincide: hence they are equal; and the surface DQF-APC. For a like reason, the surface FQE=CPB, and the surface DQE=APB; hence we

The circle which passes through the three points A, B, C, or which cir cumscribes the triangle ABC, can only be a small circle of the sphere; for if it were a great circle, the three sides AB, BC, AC, would lie in one plane, and the triangle ABC would be reduced to one of its sides.

have DQF+FQE-DQE=APC+CPB-APB, or DFE= ABC; hence the two symmetrical triangles ABC, DEF are equal in surface.

Scholium. The poles P and Q might lie within triangles ABC, DEF: in which case it would be requisite to add the three triangles DQF, FQE, DQE, together, in order to make up the triangle DEF; and in like manner, to add the three

triangles APC, CPB, APB, together, F in order to make up the triangle ABC in all other respects, the de

monstration and the result would still be the same.

B

PROPOSITION XIX. THEOREM.

If the circumferences of two great circles intersect each other on the surface of a hemisphere, the sum of the opposite triangles thus formed, is equivalent to the surface of a lune whose angle is equal to the angle formed by the circles.

Let the circumferences AOB, COD, intersect on the hemisphere OACBD; then will the opposite triangles AOC, BOD, be equal to the lune whose angle is BOD.

N

D

B

For, producing the arcs OB, OD, on the other hemisphere, till they meet in N, the arc OBN will be a semi-circumference, and AOB one also; and taking OB from each, we shall have BN=AO. For a like reason, we have DN=CO, and BD=AC. Hence, the two triangles AOC, BDN, have their three sides respectively equal; they are therefore symmetrical; hence they are equal in surface (Prop. XVIII.): but the sum of the triangles BDN, BOD, is equivalent to the lune OBNDO, whose angle is BOD: hence, AOC+BOD is equivalent to the lune whose angle is BOD.

Scholium. It is likewise evident that the two spherical pyramids, which have the triangles AOC, BOD, for bases, are together equivalent to the spherical ungula whose angle is BOD.