part EQ by a plane PQ parallel to NF, equal to the cylinder ABCD. In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the parallelogram PH (Cor. 1. 4. 1. Sup.), and cut off from the parallelogram EH, a part OR equal to the polygon AGKBLM. The point R will fall between P and N. On the polygon AGKBLM let an upright prism AGBCD be constituted of the same altitude with the cylinder, which will therefore be less than the cylinder, because it is within it (16. 3. Sup.); and if through the point. R a plane RS parallel to NF be made to pass, it will cut off the parallelopiped ES equal (2. Cor. 8. 3. Sup.) to the prism AGBC, because its base is equal to that of the prism, and its altitude is the same. But the prism AGBC is less than the cylinder ABCD, and the cylinder ABCD is equal to the parallelopiped EQ, by hypothesis; therefore, ES is less than EQ, and it is also greater, which is impossible. The cylinder ABCD, therefore, is not less than the parallelopiped EF; and in the same manner, it may be shewn not to be greater than EF.

PROP. XVIII. THEOR.

If a cone and cylinder have the same base and the same altitude, the cone is the third part of the cylinder.

Let the cone ABCD, and the cylinder BFKG have the same base, viz. the circle BCD, and the same altitude, viz. the perpendicular from the point A upon the plane BCD, the cone ABCD is the third part of the cylinder BFKG.

If not, let the cone ABCD be the third part of another cylinder LMNO, having the same altitude with the cylinder BFKG, but let the bases BCD and LIM be unequal; and first, let BCD be greater than LIM.

Then, because the circle BCD is greater than the circle LIM, a polygon may be inscribed in BCD, that shall differ from it less than LIM does (4. 1. Sup.), and which, therefore, will be greater than LIM. Let this be the polygon BECFD; and upon BECFD, let there be constituted the pyramid ABECFD, and the prism BCFKHG.

Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG is greater than the cylinder LMNO, for they have the same altitude, but the prism has the greater base. But the pyramid ABECFD is the third part of the prism (15. 3. Sup.) BCFKHG, therefore it is greater than the third part of the cylinder LMNO. Now, the cone ABECFD is, by hypothesis, the third part of the cylinder LMNO, therefore the pyramid ABECFD is greater than the cone ABCD, and it is also less, because it is inscribed in the cone, which is impossible. Therefore, the cone ABCD is not less than the third part of the cylinder BFKG: And in the same manner, by circumscribing a polygon about the circle BCD, it may be shewn that the cone ABCD is not greater than the third part of the cylinder BFKG; therefore. it is equal to the third part of that cylinder.

may

PROP. XIX. THEOR.

If a hemisphere and a cone have equal bases and altitudes, a series of cylinders be inscribed in the hemisphere, and another series may be described about the cone, having all the same altitudes with one another, and such that their sum shall differ from the sum of the hemisphere, and the cone, by a solid less than any given solid.

Let ADB be a semicircle of which the centre is C, and let CD be at right angles to AB; let DB and DA be squares described on DC, draw CE, and let the figure thus constructed revolve about DC: then, the sector BCD, which is the half of the semicircle ADB, will describe a hemisphere having C for its centre (7 def. 3. Sup.), and the triangle CDE will describe a cone, having its vertex to C, and having for its base the circle (11. def. 3. Sup.) described by DE, equal to that described by BC, which is the base of the hemisphere. Let W be any given solid. A series of cylinders may be inscribed in the hemisphere ADB, and another described about the cone ECI, so that their sum shall differ from the sum of the hemisphere and the cone, by a solid less than the solid W.

Upon the base of the hemisphere let a cylinder be constituted equal to W, and let its altitude be CX. Divide CD into such a number of equal parts, that each of them shall be less than CX; let these be CH, HG, GF, and FD. Through the points F, G, H, draw FN, GO, HP parallel to CB, meeting the circle in the points K, L and M; and the straight line CE in the points Q, R and S. From the points K, L, M draw Kf, Lg, Mh, perpendicular to GO, HP and CB; and from Q, R, and S, draw Qq, Rr, Ss, perpendicular to the same lines. It is evident, that the figure being thus constructed, if the whole revolve about CD, the rectangles Ff, Gg, Hh will describe cylinders (14. def. 3. Sup.) that will be circumscribed by the hemispheres BDA; and the rectangles DN, Fq, Gr, Hs, will also describe cylinders that will circumscribe the cone ICE. Now, it may be demonstrated, as was done of the prisms inscribed in a pyramid (13. 3. Sup.), that the sum of all the cylinders described within the hemisphere, is exceeded by the hemisphere by a solid less than the cylinder generated by the rectangle HB, that is, by a solid less than W, for the cylinder generated by HB is less than W. In the same manner, it may be demonstrated, that the sum of the cylinders circumscribing the cone ICE is greater than

the cone by a solid less than the cylinder generated by the rectangle DN, that is, by a solid less than W. Therefore, since the sum of the cylinders inscribed in the hemisphere, together with a solid less than W, is equal to the hemisphere; and, since the sum of the cylinders described about the cone is equal to the cone together with a solid less than W; adding equals to equals, the sum of all these cylinders, together with a solid less than W, is equal to the sum of the hemisphere and the cone together with a solid less than W. Therefore, the difference between the whole of the cylinders and the sum of the hemisphere and the cone, is equal to the difference of two solids, which are each of them less than W; but this difference must also be less than W, therefore the difference between the two series of cylinders and the sum of the hemisphere and cone is less than the given solid W.

PROP. XX. THEOR.

The same things being supposed as in the last proposition, the sum of all the cylinders inscribed in the hemisphere, and described about the cone, is equal to a cylinder, having the same base and altitude with the hemisphere.

Let the figure BCD be constructed as before, and supposed to revolve about CD; the cylinders inscribed in the hemisphere, that is, the cylinders described by the revolution of the rectangles Hh, Gg, Ff, together with those described about the cone, that is, the cylinders described by the revolution of the rectangles Hs, Gr, Fq, and DN are equal to the cylinder described by the revolution of the rectangle BD.

Let L be the point in which GO meets the circle ABD, then, because CGI, is a right angle if CL be joined, the circles described with the distances CG and GL are equal to the circle described with the distance CL (2. Cor. 6. 1 Sup.) or GO; now, CG is equal to GR, because CD is equal to DE, and therefore also, the circles described with the distance GR and GL are together equal to the circle described with the distance GO, that is, the circles described by the revolution of GR and GL about the point G, are together equal to the circle described by the revolution of GO about the same point G; therefore also, the cylinders that stand upon the two first of these circles, having the common altitudes GH, are equal to the

cylinder which stands on the remaining circle, and which has the same altitude GH. The cylinders described by the revolution of the rectangles Gg, and Gr are therefore equal to the cylinder described by the rectangle GP. And as the same may be shewn of all the rest, therefore the cylinders described by the rectangles Hh, Gg, Ff, and by the rectangles Hs, Gr, Fq, DN, are together equal to the cylinder described by BD, that is, to the cylinder having the same base and altitude with the hemisphere.

PROP. XXI. THEOR.

Every sphere is two-thirds of the circumscribing cylinder.

Let the figure be constructed as in the two last propositions, and if the hemisphere described by BDC be not equal to two-thirds of the cylinder described by BD, let it be greater by the solid W. Then, as the cone described by CDE is one-third of the cylinder (18. 3. Sup.) described by BD, the cone and the hemisphere together will exceed the cylinder by W. But

that cylinder is equal to the sum of all the cylinders described by the rectangles Hh, Gg, Ff, Hs, Gr, Fq, DN (20. 3. Sup.); therefore the hemisphere and the cone added together exceed the sum of all these cylinders by the given solid W, which is absurd; for it has been shewn (19. 3. Sup.), that the hemisphere and the cone together differ from the sum of the cylinders by a solid less than W. The hemisphere is therefore equal to two-thirds of the cylinder described by the rectangle BD; and therefore the whole sphere is equal to two-thirds of the cylinder described by twice the rectangle BD, that is, to two-thirds of the circumscribing cylinder.

END OF THE SUPPLEMENT TO THE ELEMENTS.

ELEMENTS

OF

PLANE TRIGONOMETRY.

TRIGONOMETRY is the application of Arithmetic to Geometry: or, more precisely, it is the application of number to express the relations of the sides and angles of triangles to one another. It therefore necessarily supposes the elementary operations of arithmetic to be understood, and it borrows from that science several of the signs or characters which peculiarly belong to it.

The elements of Plane Trigonometry, as laid down here, are divided into three sections: the first explains the principles; the second delivers the rules of calculation; the third contains the construction of trigonometrical tables, together with the investigation of some theorems, useful for extending trigonometry to the solution of the more difficult problems.

SECTION I.

LEMMA I.

An angle at the centre of a circle is to four right angles as the arc on which it stands is to the whole circumference.

Let ABC be an angle at the centre of the circle ACF, standing on the circumference AC: the angle ABC is to four right angles as the arc AC to the whole circumference ACF.

Produce AB till it meet the circle in E, and draw DBF perpendicular to AE.

Then, because ABC, ABD are two angles at the centre of the circle ACF, the angle ABC is to the angle ABD as the arc AC to the arc AD, (33. 6.) ; and therefore also, the angle ABC is to four times the angle ABD as the arc AC to four times the arc AD (4. 5.).

But ABD is a right angle, and therefore four times the arc AD is equal to

H

E

K

B

GA