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Let BCD be a triangle, and the arch CA perpendicular to the base BD; the co-fine of the angle B will be to the co-fine of the angle D, as the line of the angle BCA to the fine of the angle DCA.

For by 22. the co-fine of the angle B is to the fine of the angle BCA as the co-Gne of the Gide AC is to the radius ; that is, by Prop. 22. as) the co-Gne of the angle D to the line of the angle DCA; and, by permutation, the co-line of the angle B is to the co-line of the angle D, as the line of the angle BCA to the line of the angle DCA. Q. E. D.

PRO P. XXV. Fig. 17. 18.

TH

"HE same things remaining, the co-lines of the fides

BC, CD, are proportional to the co-fines of the bases BA, Ab.

For by 21. the co-line of BC is to the co-line of BA, as (the co-line of AC to the radius ; that is, by 21. as) the co-line of CD is to the co-fine of AD: Wherefore, by permutation, the co-lines of the sides BC, CD are proportional to the colines of the bases BA, AD. Q. E. D.

PRO P. XXVI. Fig. 17. 18.

HE fame construction remaining, the fines of the

bases BA, AD are reciprocally proportional to the tangents of the angles B and D at the base.

THE

For by 17. the line of BA is to the radius, as the tangent of AC to the tangent of the angle B; and by 17. and inversion, the radius is to the fine of AD, as the tangent of D to the tangent of AC: Therefore, ex aequo perturbate, the line of BA is to the line of AD, as the tangent of D to the tangent of B.

PROP.

PRO P. XXVII. Fig. 17. 18.

THE Co-sines of the vertical angles are reciprocally

proportional to the tangents of the sides.

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For by prop. 20. the co-one of the angle BCA, is to the ra. dius as the tangent of CA is to the tangent of BC; and by the fame prop. 20. and by inversion, the radius is to the co-Gine of the angle DCA, as the tangent of DC to the tangent of CA : Therefore, ex aequo perturbate, the co- line of the angle BCA is to the co-line of the angle DCA, as the tangent of DC is to the tangent of BC. Q. E. D.

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IN right-angled plain triangles, the hypothenuse is to

the radius, as the excess of the hypothenuse above either of the sides to the versed fine of the acute angle adjacent to that side, or as the sum of the hypothenuse, and either of the sides to the versed sine of the exterior angle of the triangle.

Let the triangle ABC have a right angle at B; AC will be to the radius as the excess of AC above AB, to the versed line of the angle A adjacent to AB; or as the sum of AC, AB to the versed fine of the exterior angle CAK.

With any radius DE, let a circle be described, and from D the centre let DF be drawn to the circumference, making the angle EDF equal to the angle BAC, and from the point F, let FG be drawn perpendicular to DE: Let AH, AK be made equal to AC, and DL to DE: DG therefore is the co-fine of the angle EDF or BAC, and GE its versed line: And because of the equiangular triangles ACB, DFG, AC or AH is to DF or DE, as AB to DG: Therefore (19. 5.) AC is to the radius DE as BH to GE, the versed line of the angle EDF or BAC: And since AH is to DE, as AB to DG, (12. 5.) AH or AC will be to the radius DE as KB to LG, the verled line of the angle LDF or KAC. Q. E. D.

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PRO P. XXVIII. Fig. 21. 22.

IN

N any spherical triangle, the rectangle contained by

the lines of two sides, is to the square of the radius, as the excess of the versed lines of the third fide or base, and the arch, which is the excess of the sides, is to the versed sine of the angle opposite to the base.

Let ABC be a spherical triangle, the rectangle contained by the lines of AB, BC will be to the square of the radius, as the excess of the versed lines of the base Ac, and of the arch, which is the excels of AB, BC to the versed sine of the angle ABC opposite to the base.

Let D be the centre of the sphere, and let AD, BD, CD be joined, and let the fines AE, CF, CG of the arches AB, BC, AC be drawn ; let the fide BC be greater than BA, and let BH be made equal to BC; AH will therefore be the excess of the fides BC, BA; let HK be drawn perpendicular to AD, and fince AG is the versed fine of the base AC, and AK the verled One of the arch AH, KG is the excess of the versed lines of the base AC, and of the arch AH, which is the excess of the fides BC, BA: Let GL likewise be drawn parallel to KH, and let it meet FH in L, let CL, DH be joined; and let AD, FH meet each other in M.

Since tberefore in the triangles CDF, HDF, DC, DH are equal, DF is common, and the angle FDC equal to the angle FDH, because of the equal arches BC, BH, the base HF will be equal to the base FC, and the angle HFD equal to the right angle CFD: The straight line DF therefore (4. 11.) is at right angles to the planc CFH: Wherefore the plane CFH is at right angles to the plane BDH, which paffes through DF. (18. 11.) In like manner, Gnce DG is at right angles to both GC and GL, DG will be perpendicular to the plane CGL ; therefore the plane CGL is at right angles to the plane BDH, which paffes through DG: And it was fhown, that the plane CFH or CFL, was perpendicular to the fame plane BDH ; therefore the common section of the planes CFL, CGL, viz. the straight line GL, is perpendicular to the plane BDA, (19. 11.) and therefore CLF is a right angle: In the triangle CFL having the right angles CLF, by the lemma CF

is to the radius as LH, the excess, viz. of CF or FH above FL, is to the versed fine of the angle CFL ; but the angle CFL is the inclination of the planes BCD, BAD, since FC, FI, are drawn in them at right angles to the common section BF: The spherical angle ABC is therefore the same with the angle CFL; and therefore CF is to the radius as LH to the versed fine of the spherical angle ABC ; and since the triangle AED is equiangular (to the triangle MFD, and therefore) to the triangle MGL, AE will be to the radius of the sphere AD, (as MG to ML; that is, because of the parallels as) GK to LH: The ratio therefore which is compounded of the ratios of AE to the radius, and of CF to the same radius ; that is, (23. 6.) the ratio of the rectangle contained by AE, CF to the square of the radius, is the same with the ratio compounded of the ratio of GK to LH, and the ratio of LH to the versed line of the angle ABC ; that is, the same with the ratio of GK to the verfed fine of the angle ABC; therefore, the rectangle contained by AE, CF, the lines of the fides AB, BC, is to the square of the radius as GK, the excess of the versed lines AG, AK, of the base AC, and the arch AH, which is the excess of the sides to the versed fine of the angle ABC opposite to the base AC. Q. E. D.

PRO P. XXIX. Fig. 23.

THE rectangle contained by half of the radius, and

the excess of the versed fines of two arches, is equal to the rectangle contained by the fines of half the fuin, and half the difference of the same arches.

I ec AB, AC be any two arches, and let AD be made equal to AC the less; the arch DB therefore is the sum, and the arch CB the difference of AC, AB: Through E the centre of the circle, let there be drawn a diameter DEF, and AE joined, and CD likewise perpendicular to it in G, and let BH be perpendicular to AE, and AH will be the versed line of the arch AB, and AG the versed Gne of AC, and HG the excess of these verfed fines : Let BD, BC, BF be joined, and FC also meeting BH in K.

Since therefore BH, CG are parallel, the alternate angles BKC, KCG will be equal; but KCG is in a semicircle, and K k 2

therefore

therefore a right angle ; therefore BKC is a right angle ; and in the triangles DFB, CBK, the angles FDB, BCK in the same segment are equal, and FBD, BKC are right angles ; the triangles DFB, CBK are therefore equiangular, wherefore DF is to DB, as DC to CK, or HG; and therefore the rectangle contained by the diameter DF, and HG is equal to that contained by DB, BC; wherefore the rectangle contained by a fourth part of the diameter, and HG, is equal to that contained by the halves of DB, BC: But half the chord DB is the fine of half the arch DAB, that is, half the sum of the arches AB, AC; and half the chord of BC is the line of half the arch BC, which is the difference of AB, AC. Whence the proposition is manifeft.

PRO P. XXX. FIG. 19. 24:

HE rectangle contained by half of the radius, and

the versed fine of any arch, is equal to the square of the line of half the same arch.

T.

Let AB be an arch of a circle, C its centre, and AC, CB, BA being joined: Let AB be bisected in D, and let CD be joined, which will be perpendicular to BA, and bisect it in E. (4. 1.) BE or AE therefore is the line of the arch DB or AD, the half of AB : Let BF be perpendicular to AC, and AF will be the versed line of the arch BA ; but, because of the limilar triangles CAE, BAF, CA is to AE as AB, that is, twice AE to AF; and by halving the antecedents, half of the radius CA is 10 AE the line of the arch AD, as the same AE to AF the versed one of the arch AB. Wherefore by 16. 6. the proposition is manifest.

PRO P. XXXI. Fig. 25,

Na spherical triangle, the rectangle contained by the

sines of the two sides, is to the square of the radius, as the rectangle contained by the fine of the arch which is half the sum of the base, and the excess of the sides, and the fine of the arch, which is half the difference of the fame to the square of the fine of half the angle oppofite to the base.

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