under the tangents of the half sum and half difference of AD and BD (1. 6 and 11. 5 Eu.); but the first and third terms of these proportionals are equal, being each equal to the square of radius (Cor. 4 Def. Pl. Tr. and 16. 6 Eu.), therefore the second and fourth terms are equal (14. 5 Eu., namely, the rectangles under the tangents of the half sum and half difference of AC and BC, and under the tangents of the half sum and half difference of AD and BD.

PROP. XXIX. THEOR.

The same things being supposed, the rectangle under the tangents of the half sum and half difference of the angles at the base, is equal to the rectangle under the cotangent of the half sum and tangent of the half difference of the segments of the vertical angle, made by the legs with the perpendicular.

The same figure and construction remaining as in the preceding proposition, the rectangle under the tangents of the half sum and half difference of the angles A and ABC, is equal to the rectangle under the cotangent of the half sum and tangent of the half difference of the angles ACD and BCD.

For the cosine of the angle A is to the cosine of the angle ABC, as the sine of the angle ACD is to the sine of the angle BCD (Cor. 1. 26 Sph. Tr.); therefore, by comparing the sums and differences of the terms, the sum of the cosines of the angles A and ABC is to their difference, as the sum of the sines of ACD and BCD is to their difference (17. 18 and 22. 5 Eu.); but the sum of the cosines of the angles A and ABC is to their difference, as the cotangent of half their sum is to the tangent of half their difference (5 Pl. Tr.), and the sum of the sines of the angles ACD and BCD is to their difference, as the tangent of half their sum is to the tangent of half their difference (4 Pl. Tr); therefore, substituting the latter ratios for the former, the cotangent of the half sum of the angles A and ABC is to the tangent of half their difference, as the tangent of the half sum of the angles ACD and BCD is to the tangent of half their difference; and, forming rectangles from the two first terms, with a common side of the tangent of the half sum of the angles A and ABC, and from the two last, with a common side of the cotangent of the half sum of the angles ACD and BCD, the rectangle under the tangent and cotangent of the half sum of the angles A and ABC, is to the

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rectangle under the tangents of their half sum and half difference, as the rectangle under the tangent and cotangent of the half sum of the angles ACD and BCD, is to the rectangle under the cotangent of the half sum of these angles and the tangent of half their difference (1. 6 and 11. 5 Eu.); but the first and third of these four proportionals are equal, being each equal to the square of radius (Cor. 4 Def. Pl. Tr. and 16. 6 Eu.), there fore the second and fourth terms are equal (14. 5 Eu.), namely, the rectangles under the tangents of the half sum and half difference of the angles A and ABC, and under the cotangent of the half sum and tangent of the half difference of the angles ACD and BCD.

LEMMA I.

The rectangle under half the radius and the difference of the versed sines of any two arches, is equal to the rectangle under the sines of half the sum and half the difference of the same arches.

D

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B

M

HLK A

Let AB and AD be two unequal arches, and let their difference BD be bisected in E; AE is equal to half the sum, and DE to half the difference of those arches. Let C be the centre of the circle, and let ABF be taken equal to a quadrant; join CF, CE, CB, CA and DB, draw DH, EL and BK at right angles to CA, BG to DH, and DM to CB, and let DB meet CE in I. GB or HK is the difference of their versed sines HA and KA, and DI is the sine of half their difference. And since the triangles CLE and DGB are equiangular, because of the right angles at L and G, and the angle BDG at the circumference insisting on an arch equal to the two arches AB and AD taken together, and of course double to that AE which subtends the angle ECL at the centre, EL is to GB or HK, as CE is to DB (4.6 and 16. 5 Eu.), or (15. 5 Eu.), as the half of CE or the half of radius. is to the half of DB, or, which is equal, to DI; therefore the rectangle under half the radius, and HK the difference of the versed sines of the arches AB and AD, is equal to the rectangle under EL the sine of the half sum, and DI the sine of the half difference of the same arches (16, 6 Eu).

LEMMA II.

The rectangle under half the radins, and the versed sine (MB, see fig. to prec. prop.), or any arch (BD), is equal to the square of the sine (BI) of half the same arch.

For the triangles CBI and DMB are, because of the right angles at I and M, and the common angle at B, equiangular.Therefore MB is to BD, as BI is to BC or radius (4. 6 Eu.), and, halving the consequents, MB is to the half of BD or to BI, as BI is to the half of radius (Theor. 2. 15. 5 Eu.); therefore the rectangle under half the radius and MB, is equal to the square of BI (17.6 Eu).

PROP XXX. THEOR.

The rectangle under the sines of the legs of any spherical triangle, is to the rectangle under the sines of the excesses of the half sum of all the sides above each of the legs, in a duplicate ratio of radius to the sine of half the vertical angle.

Let AC be the greater of

On CA

the legs AC and BC, and D the centre of the sphere. produced, take CP equal to a quadrant, and on CB produced, take CM equal to CA, and CN equal to a quadrant. From the pole C let the arch of a great circle PKN and of the less one A&M be described, join DC, DB, DN and DP; let FM be the common section of the planes DCN and AQM, which let DB meet in O, and join FA; let fall the perpendiculars PG and AH on DN and FM, and BE on DC; let fall the perpendiculars HI and ML on DB, and join AI.

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Because the plains DPN and FAM are perpendicular to the right line DC drawn from their pole C to the centre D of the sphere (Cor. 2 and 4. 2 Sph. Tr. and 19. 2 Sup.), they are perpendicular to the plain DCN (9. 2 Sup.); and because the plain DPN is perpendicular to the plain DCN, and PG perpendicular to their common section DN (Constr.), PG is perpendicular to a right line drawn from G in the plain DCN perpendicular to DN (Def. 4 and 3. 2 Sup.), and therefore to the plain DCN (2. 2 Sup.); in like manner, AH may be proved to be perpendicular to the plain DCN; whence, HI being perpendicular to DB (Constr.), DB is perpendicular to IA (Cor. 9. 2 Sup.); therefore IB is the versed sine of the arch AB, as LB is of the arch MB, the semidiameter of the sphere being radius (Def. 7. Pl. 1r).

And because the arches AM and PN are circular, having the centres at F and D in the right line DC (19. 2 Sup.), the triangles FAM and DPN are isosceles, and, because their vertical angles at F and D are equal (15 and 12. 2 Sup.), equiangular; and AH being perpendicular to FM and PG to DN (Constr.), the triangles into which these perpendiculars divide them are similar, therefore FM is to DN, as HM is to GN (4. 6 and 16 and 22. 5 Eu.); and, BE being drawn at right angles to DC, because the triangles DEB, HIO and MLO are equiangular, EB is to DB, as IL is to HM (4. 6 and 19. 5 Eu.), and compounding these two ratios, the rectangle under FM the sine of CM or CA, and EB the sine of CB, is to the rectangle under DN and DB, or the square of radius, as the rectangle under HM and IL is to the rectangle under HM and GN (23. 6 and 22. 5 Eu.), or, which is equal (1. 6 Eu.), as IL is to GN, or, which is equal (by the same), as the rectangle under the half of radius and IL, is to the rectangle under the half of radius and GN.

But the rectangle under the half of radius and IL, the difference of the versed sines of the arches AB and BM, is equal to the rectangle under the sines of the half sum and half difference of the arches AB and BM (Lem. 1 to this prop); and because CA, is equal to CM, the half sum of AB and BM is equal to the excess of the half sum of AB and AC above the half of BC, or, to the excess of the half sum of the sides AB, AC and BC above BC; and the half difference of AB and BM is equal to the excess of the half sum of AB and BC above the half of AC, or, to the excess of the half sum of the sides AB, BC and AC above AC; also the rectangle under the half of radius and GN, the versed sine of the arch PN, is equal to the square of the sine of half the same

arch PN (Lem. 2 to this prop.), or, of half the angle ACB, of which PN is the measure (3 Sph, Tr.); therefore the rectangle under the sines of CA and CB, is to the square of radius, as the rectangle under the sines of the excesses of the half sum of all the sides AB, BC and AC above each of the legs AC and BC, is to the square of the sine of half the vertical angle ACB, and by alternating, the rectangle under the sines of AC and CB, is to the rectangle under the sines of the excesses of the half sum of AB, BC and AC above AC, and of the same half sum above BC, as the square of radius, is to the square of the sine of half the angle ACB (16,5 Eu.); or, which is equal (20. 6 Eu.), in a duplicate ratio of radius to the sine of half the angle ACB.