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16. In any plane triangle it will be, as the base, to the sum of the two other sides, so is the difference of those sides, to the difference of the segments of the base made by a perpendicular let fall from the vertical angle.
From the centre c (namely, the vertical angle of the triangle ABc) with the distance of the greater side Ac describe the circumference of a circle, meeting AB, CB, produced, in the points E, F, G. Then, it is obvious that GB is equal to the sum of the sides, AC, CB, and FB equal to their difference. And because cD is perpendicular to AE, AD is equal to DE, (Euc. iii. 3). Wherefore, of the two AB, BE, one is the sum of the segments of the base DA, DB, and the other their difference. But, (Euc. iii. 35) the rectangle under GB, and BF, is equal to the rectangle under AB and BE. Consequently (Euc. vi. 16) AB : GB :: BF : BE; which agrees with the proposition.
17. In any plane triangle it will be, as twice the rectangle under any two sides, is to the difference of the sum of the squares of those two sides and the square of the base, so is the radius to the cosine of the angle contained by the two sides.
Let ABC be a plane triangle. From A, C one extremity of the base AB, draw AD per- Yo pendicular to the opposite side Bc: then (Euc. ii. 12, 13) the difference of the sum of the squares of Ac, cB, and the square of A B
the base AB, is equal to twice the rect- D angle Bc. CD. But twice the rectangle C Bc. CA, is to twice the rectangle BC. CD, /*s that is, to the difference of the sum of the squares of Ac, BC, and the square of AB, A. B as cA to cD; that is, (prop. 13) as radius to the sine of cAD, or, as radius to the cosine of AcB. Cor. When unity is assumed as radius, we have Ac” + BC” – AB”
18. As the sum of the tangents of any two unequal angles, is to their difference, so is the sine of the sum of those angles, to the sine of their difference. Let BCA, ACD, be the two proposed C angles, BA and AD their tangents to the radius CA. Take 1A = AD, join E% c1, and draw DE, de, IK, ik, perpendicular to Bc. Then, it is evident, since K% d! IA = AD, and CA perpendicular to BD, that c1 = CD, ICA = DcA, and there- B I A D fore BCI = BCA – AcD: also that ed is the sine of the sum of the given angles, and ik the sine of their differe ce,to the assumed radius Ac. Now, by reason of the similar triangles BDE, BIK, it will be, BD (= BA + AD): e1 (= BA – AD):: DE: IK. Also, because the triangles coe, Cde, are similar, as well as the triangles CIK, cik; we have CD: cal:: DE : de, and cI (= cD): ci (= ca) :: Ik: ik; therefore DE: Ik :: de : ik. And consequently, BD: BI:: de : ik; that is, tan BCA + tan DcA: tan BCA – tan DCA :: sin (BCA + DCA): sin (BCA - DCA). Cor. Hence it follows, that the base of a plane triangle BCD, is to the difference of its two segments BA,
AD, as the sine of the whole angle at the vertex, to the sine of the difference of the angles at the base.
19. As the sine of the difference of any two unequal angles, is to the difference of their sines, so is the sum of those sines, to the sine of the sum of the angles. Let A and B be the two unequal an- C gles; both being bounded on one side by the line AB, and on the other by lines, which, produced, meet at c, and form a rectilinear triangle AcB. Demit cd per- - A D a B pendicularly from c upon AB; make Da = DA, and join ca. Then we shall have a B = DB – DA, Ca = CA, Cap = CAD, cab = supp. CAB; and conseq. (ch. i. 19 c) sin cAB = sin cab. Also (by the ...) sin c = sin (A + B); and acB = Aac – B = A — B. Hence, in A AcB, sin C, or sin (A + B): sin A :: AB : cB. Also, in A acB, sin C, or sin (A – B); sin A :: a B : cB. Therefore, ab: AB :: sin (A – B): sin (A + B). Again (prop. 14) cB : CA :: sin A : sin B. ... comp. and div. CB – CA : CB + c A :: sin A — sin B : sin A + sin B. But, by the inversion of prop. 16, - aB : CB - CA :: CB + CA: AB. Therefore, comparing the corresponding antecedents and consequents, in the 3d and 5th analogies, there re'sults, . sin (A – B): sin A — sin B:: sin A + sin B : sin (A + B). Cor. If A and B be to each other as n + 1 to n, then the last proportion will become sin A : sin (n + 1) A — sin na :: sin (n + 1) A + sin na : sin (2n + 1) A.
20. Of the propositions in this chapter, some are useful in the solutions of plane triangles, and will find their applications in the next chapter; others are useful in the construction of tables of natural sinés, tangents, &c. To glance, for a moment, at this application, in passing, suppose it were required to find the natural sines to each of the 10 first minutes of the quadrant. The radius of a circle being 1, the semi-circumference is known to be 3.14159265358979. This being divided successively by 180 and 60, or at once by 10800, gives •00029088820.86657, for the arc of 1 minute. Of so small an arc the sine, chord, and arc, differ almost imperceptibly from the ratio of equality; so that the first ten of the preceding figures, that is, 0002908882 may be regarded as the sine of 1*; and in fact the sine given in the tables which run to seven places of figures is ‘0002909. By chap. i. art. 19, we have, for any arc, cos = V (1 — sin”). This theorem gives, in the present case, cos 1 = .9999999577. Then, by prop. 12, cor. 2, (A), of this chapter, we shall have 2 cos 1’ x sin 1' – sin O' = sin 2' = .0005817764, 2 cos 1’ x sin 2" – sin 17 = sin 3’ = 0008726646 2 cos 1’ x sin 3 – sin 2' = sin 4 = -0011635.526 2 cos 1’ x sin 4' — sin 3' = sin 5’ = -0014544.407 2 cos 1’ x sin 5’ — sin 4' = sin 6 = -0017453284 &c. &c. &c. Thus may the work be continued to any extent, the whole difficulty consisting in the multiplication of each successive result by the quantity 2cos 1'=1-9999999.154. Or, the sines of 1° and 2 being determined, the work might be continued by the last proposition, thus: sim 1": sin 2" — sin 1":: sin 2' + sin 1": sin 3’ sin 2 : sin 3’ — sin 17 :: sin 3' + sín 1": sin 4' sin 3’: sin 4' — sin 1":: sin 4' + sin 1": sin 5’ sin 4' : sin 5' — sin 1":: sin 5' + sín 1": sin 6' &c. &c. &c. In like manner, the computer might proceed for the sines of degrees, &c. thus: sin 19 : sin 29 – sin 19:: sin 22 + sin 19 : sin 3° sin 22 : sin 39 — sin 19:: sin 30 + sin 19 : sin 4° sin 3°: sin 4° – sin 19 :: sin 4° 4- sin 12 : sin 5° &c. &c. &c.
To check and verify operations like these, the proportions should be changed at certain stages. Thus, sin 19 : sin 3° – sin 2°:: sin 3° -- sin 29: sin. 5° sin-1°: sin 4° — sin 3°:: sin 4° -- sin 3°: sin 7° sin 4°: sin 7° — sin 3°:: sin 7° -- sin 3°: sin 10°. The coincidence of the results thus obtained, with the analogous results in the preceding operations, will mamifestly establish the correctness of both. The sines and cosines of the degrees and minutes up to 30°, being determined by these or other processes (some of which will be indicated in the 4th chapter), they may be continued thus: sin 30° 1' = cos 1’ – sin 29°59' sin 30°2' = cos 2" – sin 29° 58' sin 30°3′ = cos 3' — sin 29°57'. And these being continued to 60°, the cosines also become known to 60°; because cos 30° 1' = sin 59° 59' cos 30°2' = sin 59° 58'. The sines and cosines from 60° to 90°, are deduced from those between 0° and 30°. For sin 60° 1' = cos 29° 59' sin 60°2' = cos 29° 58' &c. &c. &c. 21. The sines and cosines being found, the versed sines are determined by subtracting the cosines from radius, in arcs less than 90°, and by adding the cosines to radius in arcs greater than 90°. 22. The tangents may be found from the sines and
cosines. For since tan = *, (chap. i. 19),
&c. &c. &c. Above 45° the process may be considerably simplified