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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 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.

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 are 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 taken for unity, the surface of the sphere will be represented by 8. This granted, if the right angle be assumed equal to 1, the surface of the lune whose angle is A will be expressed by 2A; for,

48A : 2A.

Thus we have here two different unities; one for angles, being the right angle; the other for surfaces, being the tri-rectangular spherical triangle, or the triangle whose angles are all right angles, and whose sides are quadrants.

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 planes which bound them.


Two symmetrical spherical triangles are equivalent.

Let ABC, DEF, be two symmetrical triangles, that is to say, two triangles having their sides AB-DE, AC=DF, CB=EF, and yet incapable of coinciding with each other: we are to show that the surface ABC is equal to the surface DEF.

Let P be the pole of the small F

circle passing through the three points

A, B, C ;* from this point draw the

[blocks in formation]

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 AB on its equal DF, the equal sides will fall on each 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 circumscribes 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.


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.


For, producing the arcs OB, OD, on the other hemisphere, till they meet in N, the arc OBN will be a semicircumference, and AOB one also; and taking OB from both, 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; besides they are so placed as to be 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.


The surface of a spherical triangle is measured by the excess of the sum of its three angles above two right angles.


Let ABC be the proposed triangle: produce its sides till they meet the great circle DEFG drawn at pleasure without the triangle. By the last Theorem, the two triangles ADE, AGH, are together equivalent to the lune whose angle is A, and which is measured by 2A (Prop. XVII. Cor. 2.). Hence we have ADE+AGH=2A; and for a like reason, BGF+BID=2B, and CIH+CFE=2C. But the sum of those six triangles exceeds the hemisphere by twice the triangle ABC, and the hemisphere is represented by 4; therefore twice the triangle ABC is equal to 2A+2B+2C-4; and consequently, once ABC=A+B+ C-2; hence every spherical triangle is measured by the sum of all its angles minus two right angles.

Cor. 1. However many right angles there be contained in this measure, just so many tri-rectangular triangles, or eighths of the sphere, will the proposed triangle contain (Prop. XVII. Cor. 2.). If the angles, for example, are each equal to 4 of a right angle, the three angles will amount to 4 right angles, and the proposed triangle will be represented by 4-2 or 2; therefore it will be equal to two tri-rectangular triangles, or to the fourth part of the whole surface of the sphere.

Cor. 2. The spherical triangie ABC is equivalent to the lune whose angle is A+B+C -1; likewise the spherical pyramid, 2 which has ABC for its base, is equivalent to the spherical ungula whose angle isA+B+C



Scholium. While the spherical triangle ABC is compared with the tri-rectangular triangle, the spherical pyramid, which has ABC for its base, is compared with the tri-rectangular pyramid, and a similar proportion is found to subsist between them. The solid angle at the vertex of the pyramid, is in like manner compared with the solid angle at the vertex of the trirectangular pyramid. These comparisons are founded on the coincidence of the corresponding parts. If the bases of the

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