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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
Two symmetrical spherical triangles are equivalent.
Let P be the pole of the small F circle passing through the three points 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.
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 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 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 demonstration and the result would still be the same.
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.
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.
PROPOSITION XX. THEOREM.
The surface of a spherical triangle is measured by the excess of the sum 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 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.T (Prop. XVII. Cor. 2.). Hence we have ADE+AGH=2A.T; and 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—4 T; 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, 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
pyramids coincide, the pyramids themselves will evidently coincide, and likewise the solid angles at their vertices. From this, some consequences are deduced.
First. Two triangular spherical pyramids are to each other as their bases: and since a polygonal pyramid may always be divided into a certain number of triangular ones, it follows that any two spherical pyramids are to each other, as the polygons which form their bases.
Second. The solid angles at the vertices of these pyramids, are also as their bases; hence, for comparing any two solid angles, we have merely to place their vertices at the centres of two equal spheres, and the solid angles will be to each other as the spherical polygons intercepted between their planes or faces.
The vertical angle of the tri-rectangular pyramid is formed by three planes at right angles to each other: this angle, which may be called a right solid angle, will serve as a very natural unit of measure for all other solid angles. If, for example, the the area of the triangle is of the tri-rectangular triangle, then the corresponding solid angle will also be of the right solid angle.
PROPOSITION XXI. THEOREM
The surface of a spherical polygon is measured by the sum of all its angles,minus two right angles multiplied by the number of sides in the polygon less two, into the tri-rectangular triangle.
From one of the vertices A, let diagonals AC, AD be drawn to all the other vertices; the polygon ABCDE will be divided into as many triangles minus two as E it has sides. But the surface of each triangle is measured by the sum of all its angles minus two right angles, into the trirectangular triangle; and the sum of the angles in all the triangles is evidently the same as that of all the angles of the polygon; hence, the surface of the polygon is equal to the sum of all its angles, diminished by twice as many right angles as it has sides less two, into the tri-rectangular triangle.
Scholium. Lets be the sum of all the angles in a spherical polygon, n the number of its sides, and T the tri-rectangular triangle; the right angle being taken for unity, the surface of the polygon will be measured by
(s—2 (n-2)) T, or (s—2 n+4) T
THE REGULAR POLYEDRONS.
A regular polyedron is one whose faces are all equal regular polygons, and whose solid angles are all equal to each other. There are five such polyedrons.
First. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of those triangles, by four, or by five: hence arise three regular bodies, the tetraedron, the octaedron, the icosaedron. No other can be formed with equilateral triangles; for six angles of such a triangle are equal to four right angles, and cannot form a solid angle (Book VI. Prop. XX.).
Secondly. If the faces are squares, their angles may be arranged by threes: hence results the hexaedron or cube. Four angles of a square are equal to four right angles, and cannot form a solid angle.
Thirdly. In fine, if the faces are regular pentagons, their angles likewise may be arranged by threes: the regular dodecaedron will result.
We can proceed no farther: three angles of a regular hexagon are equal to four right angles; three of a heptagon are greater.
Hence there can only be five regular polyedrons; three formed with equilateral triangles, one with squares, and one with pentagons.
Construction of the Tetraedron.
Let ABC be the equilateral triangle which is to form one face of the tetraedron. At the point O, the centre of this triangle, erect OS perpendicular to the A planc ABC; terminate this perpendicular in S, so that AS-AB; draw SB, SC: the pyramid S-ABC will be the tetraedron required.
For, by reason of the equal distances
OA, OB, OC, the oblique lines SA, SB, SC, are equally re