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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 tri

C angle 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 con tained eight times in the surface of a sphere.


Scholium. In all the preceding

C observations, we have supposed, in conformity with (Def. 1.) that spherical triangles have always each of their sides less than a semicircum


D ference; from which it follows that

A 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

E 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 semicir cumference 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.

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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 Suppose, in the first place, the arc

IC MN to be to the circumference MNPQ N as some one rational number is to another, as 5 to 48, for example. The cir

B cumference 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.

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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 XT 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 planes which bound them.


Two symmetrical spherical triangles are equivalent.


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Let ABC, DEF, be two symmetri

A cal triangles, that is to say, two tri

D angles having their sides AB=DE, AČ=DF, CB=EF, and yet incapable of coinciding with each other :

QP we are to show that the surface ABC

C is equal to the surface DEF.

Let P be the pole of the small T circle passing through the three points

B A, B, C ;* from this

point draw the 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.

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

* 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.

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 or-

der 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.

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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.

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

N 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.



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The surface of a spherical triangle is measured by the excess of

the såm of its three angles above two right angles, multiplied by the tri-rectangular triangle.

Let ABC be the proposed triangle: produce its sides till they meet the great circle DEFG drawn at pleasure without the trian- I gle. 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.T (Prop. XVII. Cor. 2.). Hence we have AÐEF AGH=2A.T ; and I for a like reason, BGF+BID=2B.T, and CIH +CFE=2C.T But the sum of these six triangles exceeds the hemisphere by twice the triangle ABC, and the hemisphere is represented by 4T; therefore, twice the triangle ABC is equal to 2A.T+2B.T+2C.T_4T; and consequently, once ABC=(A+B+C—2)T; hence every spherical triangle is measured by the sum of all its angles minus two right angles, multiplied by the tri-rectangular triangle.

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

Scholium. While the spherical triangle ABC is compared with the tri-rectangular triangle, the spherical pyramid, wbich bas 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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