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of the one equal to one side of the other, be taken equal to the triangle A B C or and let the equal sides be made to coin- KLM, and join K' M. Then, by the cide, so that the triangles A B C, EBC former case, the pyramids upon the (in the figure of Prop. 24.) may repre- bases ABC, K' L M' are equal to one sent the bases of the pyramids. Then another. And, because the triangle the triangles ABC, EBC, being equal K' L M' is equal to the triangle K L M, to one another, lie between the same the triangles K' M M' and K KM equal and parallel small circles, and which have the common side K' M are may be completed (as in Prop. 24.) into likewise equal; therefore, by the first the quadrilaterals ABCD, EBC F.

case, the pyramids, which have these And it may be shown, as in the same triangles for their bases, are equal to one proposition, that the triangles E A B, another; and, these being added to, or FDC have the three sides of the one taken from, the pyramid upon the base equal to the three sides of the other, K'LM, the whole or remaining pyraeach to each, and therefore may be mid upon the base K' L M' is equal to made to coincide (15.): wherefore, also, the whole or remaining pyramid upon the pyramids, which have these triangles the base KLM, that is to the pyramid for their bases, may be made to coincide, upon the base A BC. and are equal to one another. And hence, Lastly, let the bases of the pyramids as in Prop. 24., it was demonstrated, P, P' be equal polygons. Let a triangle by the addition and subtraction of the be found which is equal to one, and lunular portions of surface, that the qua- therefore also to the other of the drilateral A B C D is equal to the quadri- polygons, and let Q be the pyramid lateral EBCF, so, here, it may be demon- which has this triangle for its base. strated, by the addition and subtraction Then, because this triangle is equal to of the pyramidal solids (23.), which have the base of the pyramid P, it these lunular portions for their bases, divided into triangles which are equal that the pyramidal solids, which have for respectively to the triangles into which their bases the quadrilaterals A B C D the base of P is divisible. And it has and E BCF, are equal to one another. been already shown that pyramids which But, because the triangles A B C, CDA have equal triangles for their bases are have the three sides of the one equal to equal one another; and the sums of the three sides of the other, each to each, equals are equal; therefore, the pyrain the same order, they may be made to mids P and Q are equal to one another. coincide (15.), and therefore the pyramids In the same manner it may be shown that which have these triangles for their the pyramids P' and Q are equal to one bases may be made to coincide, and are another. Therefore P is equal to Pl. And equal to one another; and each of them it is evident that what has been shown is the half of the pyramidal solid which with regard to spherical pyramids, being has the quadrilateral A B C D for its base. derived from coincidence, may be shown And in the same manner it may be equally of their solid angles at the censhown, that each of the pyramids on the tre of the sphere. triangular bases, EBC, CFE is the Therefore, &c. half of the pyramidal solid on the quadrilateral base EBCF. Therefore,

PROP. 28. because the halves of equals are equal,

Any two spherical pyramids are to the pyramid upon the base A B C is equalto the pyramid upon the base E B C solid angles of the pyramids are to one

one another as their bases; and the Next, let the bases of the

pyramids another in the same ratio. be any equal triangles A B C, K L M.

Let P, P' be any two spherical pyra-

mids, and let B, B' be their bases: the
pyramid P shall be to the pyramid P'

as the base B to the base B'. B

For if the base B' be divided into any L M M

number of equal parts, then, because K

pyramids which stand upon equal bases

are equal to one another, the pyramid P' K

will be divided into the same number of L

equal parts by planes passing through

the arcs of division (27.); and if the base From LK or L K produced cut off L K B contain exactly, or with a remainder, equal to B A, and let the triangle K'LM' a certain number of parts equal to the



former, the pyramid P will contain faces, are represented, upon the surface exactly, or with a remainder, the same

a sphere described about the angular number of parts equal to the latter (27.): point, by the sides, angles, and surface therefore, P is to P' as B to B' (II. of a spherical triangle or polygon: and def. 7.)

whatever has been stated with regard to And the same may be said of the solid the latter may be understood likewise of angles.

the former. It is shown (for instance) Therefore, &c.

in Prop. 13. that if two solid angles, each Cor. Every spherical pyramid is equal of which is contained by three plane anto the third part of the product of its gles, have two plane angles of the one base and the radius of the sphere. equal to two plane angles of the other,

For, if the whole sphere be divided each to each, and likewise the dihedral into spherical pyramids, these pyramids angles contained by them equal to one will be to one another as their bases; another, the remaining plane and diheand, therefore, any one of them is dral angles of the one shall be equal to to the sum of all, as the base of that one

the remaining plane and dihedral angles to the sum of all the bases (II. 25. of the other, each to each. And, geneCor. 3.); that is, any pyramid is to the rally, all questions which relate to solid whole sphere as its base to the surface angles will be placed in the clearest of the sphere. Therefore, any pyramid view before us, when we contemplate is to the whole sphere as the product of only their representation and that of its base and the radius of the sphere to their parts, upon the surface of the the product of the whole surface and sphere. radius (IV. 26.); and because, in this proportion, the second term is equal to

SECTION 4.- Problems. the third part of the fourth (V. 17.), the first term is likewise equal to the third In the solution of the following propart of the third (II. 19. and II. 13.). blems, it is assumed,

1. That of any two points, which are Scholium.

given upon the surface of a sphere, the Hence it appears that every solid direct distance may be obtained; and, angle is measured by the spherical sur- 2. That from any given point as a face which is described with a given pole, with any given distance less than radius about the angular point, and in- à semicircumference, a circle may be tercepted between its planes. For it is

described upon the surface of a sphere. shown in the proposition that any two We may observe, however, that the first solid angles are to one another as these assumption is only used in Prob. 1 to surfaces.

obtain the diameter of the sphere; and, This measure bears an obvious ana- therefore, if the diameter is supposed to logy to the measure of a plane angle, as

be given, it may be dismissed as unnestated in the Scholium at Book III.

cessary. Constructions made within Prop. 13. The angular unit of the latter the sphere are excluded. measurement is the right angle ; into four of which the whole angular space

PROP. 29. Prob. 1. about any point in a plane is divided by

To find the diameter of two straight lines drawn at right angles

a given to one another; and each right angle is sphere A B C D. measured by a quadrant, or fourth part distance AB, describe the circle BCD;

From any point A, as a pole, with any of the circumference of a circle described about that point with a given radius.

i In like manner, if, through a point in space,

three planes be made to pass at right angles to one another, they will divide the whole angular space about that point into eight solid right angles, each of which (a solid angular unit) is measured by an octant, or eighth part of the surface of a sphere described about

F that point with a given radius.

Thus, then, the plane angles, dihedral in the circumference B C D take any angles, and magnitude of a solid angle, three points B, C, D ; describe the plane which has three or a greater number of triangle bed with its sides bc, cd, bd







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equal to the direct distances BC, CD, Then, because the planes of any two B D respectively; find e the centre of great circles cut one another in a diathe circumscribing circle (I11.59.), and meter of the sphere, the points A, a are join eb; from the point e draw e a per- opposite extremities of a diameter. pendicular to eb, and from the centre Therefore, &c. b, with a radius equal to the direct dis

PROP. 31. Prob. 3. tance B A describe a circle cutting ea To join two given points A,


ироп in a; produce a e to f, and from b draw the surface of a sphere. bf perpendicular to ba (I. 44.) to meet From the pole A, af in f: af shall be equal to the dia- with the distance meter of the sphere.

of a quadrant (29. For, if o be the centre of the sphere, Cor.), describe a and the diameter AOF be drawn cut- great circle, and ting the plane B C D in E, and E B be from the pole B, joined, A F will be the axis (def. 3.), with the same disand therefore (1.) the point E the centre, tance, describe a of the circle B C D, and the angle A E B great circle cutwill be a right angle. Also, EB is equal ing the former in the point P. From P to eb, because the triangles B C D, bcd as a pole, with the same distance, demay be made to coincide, and, then, the scribe the arc A B. Then, because AB points E, e being, each of them, the is the arc of a great circle (5.) described centre of the same circumscribing circle, between the points A, B, it joins those will likewise coincide. Therefore, be- two points on the surface of the sphere cause in the right-angled triangles AEB, (def. 5.). a eb, the hypotenuse AB is equal to Therefore, &c. the hypotenuse ab, and the side EB Cor. In the same manner, any spheto the side eb, the angle B A E is equal rical arc being given, the great circle to the angle ba e (I. 13.). Join AB, may be completed of which it is a part. BF, O B. Then, because O B is equal

PROP. 32. Prob. 4. to the half of A F, the angle ABF is

To bisect a given spherical arc A B. a right angle (I. 19. Cor. 4.); and,

From the pole A, with because, in the right-angled triangles the distance & B describe A BF, abf, the side A B and angle a circle, and from the pole BAF are equal respectively to the side B with the same distance ab and angle baf, the hypotenuses AF

AB or BA describe a cirand af are equal to one another (I. 5.), cle cutting the former cir

А and a f is equal to the diameter of the cle in the points C, D. sphere.

Join CD(31.) and let it cut Therefore, &c.

A B in E. AB is bisected Cor. Hence, a sphere being given, in E. the quadrant of a great circle may be See Book I. Prop. 43. found; for it is equal to the quadrant

Therefore, &c. of the circle which is described


It must here be observed that al. diameter af.

though the points C, D are determined PROP. 30. Prob. 2.

by the intersection of small circles, the Any point A being given upon the the sides of the spherical triangles in the

arcs CA, CB, CD, DA, DB, which are surface of a sphere, to find the opposite demonstration, are portions of great cirextremity of the diameter which passes cles; and the same remark applies to through that point.

some subsequent problems. From the pole A,

Cor. In the same manner, a spherical with the distance of

arc CD may be drawn which shall bia quadrant, (29.

sect any spherical arc A B at right anCor.) describe


PROP. 33. Prob. 5. spherical arc PQ (5.); and, from any

To draw an arc, which shall be pertwo points P, Q of

pendicular to a given spherical arc AB, this arc, as poles,

from a given point C in the same. with the same dis

From CB, or CB tance, describe two

produced, cut off CP great circles passing through A, and equal to a quadrant,

А. с cutting one another again in the point a. and from the pole P,









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with the distance PC, describe the spheri

PROP. 37. Prob. 9. cal arc CD. Then, because PC is drawn from the pole of a great circle

To describe a circle through three CD to the point C in that circle, PC given points A, B, C, upon the surface is at right angles to CD (5.) ; or, which of a sphere. is the same thing, C D is at right angles Join (31.) AB and to PC or AB.

AC, and bisect them Therefore, &c.

at right angles (32. PROP. 34. Prob. 6.

Cor.) with the arcs

DP, EP, which meet To draw an arc, which shall be per

one another in the pendicular to a given spherical arc AB, points P, P'. From from a given point C without it.

either of these points, From the pole C,

P, as a pole, with the with the distance of


distance PA, describe a quadrant, describe

a circle. It shall pass through the other a great circle, cut

英 two points B and C. For PB is equal to ting AB or AB pro

PA, because the triangles PDB, PDA duced in P, and from

have two sides of the one equal to two the pole P, with the same distance, de- sides of the other, each to each, and the scribe the spherical arc CD. For the included angles PDA, PDB equal to same reason as in the last problem, CD one another (13.); and in the same manis the perpendicular required.

ner it may be shown that PC, likewise, Therefore, &c.

is equal to PA. PROP. 35. Prob. 7.

Therefore, &c. To bisect a given spherical angle P, P of any given circle A B C may be

Cor. In the same manner, the poles BAC.

found. In A B take any point A

PROP. 38. Prob. 10.
B; make AC equal to
AB, and join BC; from

Through two given points A, B, and the poles B, C, with the

a third point C on the surface of a common distance BC, В,

C sphere, to describe two equal and paraldescribe two spherical

lel small circles ; the points A, B, C not arcs cutting one another

lying in the circumference of the same in D, and join A D (31.).

great circle. AD is the bisecting arc

Join A B, and draw D E bisecting it required. See Book I. Prop. 46. at right angles in the point D (32. Cor.); Therefore, &c.

find A' the opposite extremity of the

diameter which passes through A (30.); PROP. 36. Prob. 8.

join A' C (31.), and bisect it at right At a given point A, in a given arc

angles with the arc FP which meets AB, to make a spherical angle equal to

DĚ in the points P, P (32. Cor.); a given spherical angle C.

join PA, PC, and from the pole P with From the pole C, with the distance of a quadrant, describe a circle cutting the sides of the given angle in the points

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D D, E; and from the pole A, with the same distance, describe a circle B FG the distances PA, PC describe two cutting A B in B ; from BF G cut off small circles A GH and CKL: the BF equal to 1) E, and join AF (31.). The former shall likewise pass through the angle B A F is equal to the given angle point B, and they shall be the two equal C; for they are measured by equal arcs and parallel small circles required. BF, D E (6.)

For, if P A and P B be joined, they Therefore, &c.

will be equal to one another, beca use






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the triangles PDA, PDB have two

Bisect AB in H sides of the one equal to two sides of (32.), and produce the other, each to each, and the included H A to P, so that angles PDA, PDB equal to one ano- HP may be equal ther (13.). And, for the same reason, if to a quadrant ; PA'and PC are joined, PA' will be equal from the pole P, H to PC. But PÅ and P A' are together with the distance

қ equal to a semicircumference, because PB, describe the A'A' is a diameter of the sphere. There- small circle BCK, fore, PA and PC are together equal to and from the pole a semicircumference; and, consequently B, with the distance Q, describe a circle (3. Cor.), the parallel circles AGH, cutting the circle B°C K in C. Join CK L are equal to one another. Also, AC, B C. The triangle A B C shall be because PA is equal to PB, the circle the triangle required. AG H passes through the point B. For, if, with the distances P A, PH, Therefore, &c.

there are described from the pole P the PROP. 39. Prob. 11.

small circle A LM and the great circle To describe a triangle which shall be G, the circles A L M and B C K will be

HGN, the latter cutting the arc AC in equal to a given spherical polygon, and equal to one another, because the disshall have a side and adjacent angle the tances PA, PB are together equal to sume with a given side AB and adjacent twice the quadrant PH (I. ax. 9.), that angle B of the polygon.

is, to a semicircumference (3. Cor.); and First, let the given

they are parallel because they have the polygon be a quadri

same poles; and HGN is the great circle lateral A B C D.

to which they are parallel ; therefore, Through the two

AG is equal to GC (22.). But, because points A, C, and the B

P is the pole of the great circle HGN, third point D, describe two equal and PH is at right angles to H G (5.); and, parallel small circles (38.), and let the because the triangles GHA, GHB arc B C, which cuts one of them in C, have two sides of the one equal to two he produced to cut the other in E (22. sides the other, each to each, and the Cor. 2.), and join AE, AC (31.). Then, included angles GHA, G H B equal to because the triangles A CD, ACE lie

one another, G B is equal to GA or GC between the same equal and parallel (13.). Therefore, because in the isosceles small circles, they are equal to one ano- triangles G AB, GBC, the angles GBA, ther; and, therefore, the triangle ABC GBC are equal to the angles GAB, being added to each, the triangle ABE GCB respectively (11.), the whole angle is equal to the quadrilateral A B C D.

A B C of the triangle A B C is equal to Next, let A B C D E F be the given the sum of the two angles CAB, ACB; polygon. Join A C, AD. Make, as in wherefore the triangle ABC the the former case, the triangle ADG equal greatest that can be formed with the to the quadrilateral

two sides A B, B C, or the greatest that ADEF, the trian

can be formed with the given sides A B gle ACH equal to

and Q (26. Cor.). the quadrilateral A,

Therefore, &c. A C-D G, and the

PROP. 41. Prob. 13. triangle ABK equal

K to the quadrilateral

Through a given point A to describe ABCH. It is evi

a great circle, which shall touch two dent that the triangle ABK is equal to given equal and parallel small circles the polygon ABCDEF.

BCD, EFG. And, in each case, the triangle de

Find the point P which is the comscribed has the same side AB and angle

B with the given polygon.
Therefore, &c.


D PROP. 40. Prob. 12. Given two spherical arcs A B and Q, which are together less than a semicircumference; to place them so, that, with a third not given, they may contain the greatest surface possible.





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