Abbildungen der Seite
PDF
EPUB

Theorem 6. Cosa (1-tana)÷(1+tan2ļa).

1. By Art. 210, equa. 2d, cos a=cos2a-sin2 a 2. By Art. 93, cos=R2÷sec; that is, cos2a=R4÷sec2 a: also, by Art. 94, sec2=R2+tan2; that is, sec21a= R2+tana, and by substitution, cosa R÷(R2 +tan3 a) 1÷(1+tan2a)

3. By Art. 93, sin=R2÷cosec; that is, sin2a=R2÷cosec2 a: also, by Art. 94, cosec2=R2+cot2; that is, cosec2a R2+cota, and by substi. sin2a=R÷(R2+cota) 4. By Art. 220, cot=R2÷tan; that is, cot2a=R2÷tan2

=

a, and by substi. sin2a=R÷÷÷( R2 +· ( 5. Alg. 148,R4 ÷ (R2 +· (R2 +

R2

tan2a

R2

tan2a

:)

Ratan21+R2

=R4 ÷

(

tan2 a

[blocks in formation]

tan2a

[ocr errors]

=

=

tan2 a

sina

Retana+R2 tan2a+1

6. Substituting in 1st the values as found in 2d and 5th, cos a=1÷(1+tan2 1a) — tan2 a÷(tan21a+1)

7. Uniting terms, cos a=(1-tan2 1a)÷(1+tan21a).

Theorem 7. Cos a=(cot, a-tanja)+(cotla+tanja) 1. By the 6th theorem, cos a= (1-tana)=(1+tana)

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][subsumed][merged small][merged small]

4. By Art. 93, R2 ÷tan=cot=1÷tan=cot=1÷tana=

cotla

5. Substituting, cos a (cota-tanja)÷(cota+tanļa)

a=

Theorem 8. Sin a=2÷(cota+tanļa)

1. By theorem 2d, tan a=(1-cos a)÷sin a
2. Multiplying by sin a, sin a tana=1-cosa
3. Dividing by tan ja, sin a=(1-cosa)÷tan a

tana,

4. By the 7th theor., cos a=(cota- tana)÷(cota +

[blocks in formation]

cot la+tana-cot ja+tan a'

cotla+tana

[merged small][ocr errors]

(cotla+tana-cot la+tana)÷(tan la cota+tana) 6. Cancel. terms, sin a 2 tan a÷(tan la cota +tana) 7. Dividing by tan ja, sin a=2÷(cota+tana).

2

Theorem 9. Sin a=1÷(cot a-cot a).

1. By theorem 8th, sin a 2-(cota+tana)

2. By theorem 1st, tan a cota-2 cot a

3. Subst. in 1st, sin a=2÷(cot ja+cota-2 cota)
4. Uniting terms, sin a=2÷(2 cot a-2 cot a)
5. Dividing by 2, sin a=1÷(cot a-cot a).

Theorem 10. sin a=1÷(cot a+tanļa) 1. By theorem 1st, tana=cota-2 cot a 2. Transposing, tan la+2cot a cota

=

3. By theorem 9th, sin a-1÷(cota-cot a)
4. By substituting, sin a=1÷(tana+2cot a-cot a)
5. Uniting terms, sin a=1÷(tan la+cota).

Expression for the area of a triangle in terms of the sides.

Throughout the following process let a, b, c, (Fig. 23.) represent the sides, p the perpendicular CD, d the segment AD, S the area, and h half the sum of the sides.

1. By Euc. II. 13, a2=b2+c2-2cd

2. Transposing, 2cd=b2+c2-a3

3. Dividing by 2c, d=(b2+c2 — a2)÷2c

=

4. By Euclid I. 47, p2 —b2 —d2 — (b2 —b2 +c2 — a2 )2 ÷4c2 5. Clearing of fractions, 4c2p2=4b2 c2 — (b2 +c2 — a2)2 4b2c2-(b2+c2-a2)2 6. Dividing by 4c2, p2 =

4c2

7. Extract. the square root, p= √462c2

[merged small][ocr errors][merged small]

8. By Alg. 518, S= base X perpendicular; and, 9. By Substituting, S=cX √4b2 c2 — ( b2 +c2 —a2 )2 ÷ 2c =¦×¦√4b2 c2−(b2 +c2¬a2)2=‡√4b2c2 —(b2 +c2 —a2)2 10. By substi. S‡√(b+c+a)×(b+c−a)×(a+b−c)× √(a−b+c)

11. As h=(a+b+c), a+b+c=2h= the 1st factor; + c-a=(a+b+c)-2a=2h-2a=the 2d factor; a+b-c=

(a+b+c)-2c=2h-2c=the 3d factor; a-b+c=(a+b+c) -2b=2h-2b=the 4th factor: Hence,

12. By substi. S=√2h×2(h-a)×2(h-c)×2(h—b) 13. Multiplying, S=‡√16×h×(h—a)×(h—b)×(h−c) 14. Removing a factor (Alg. 271.) S=4×4√h×(h—a)× √(h—b)×(h—c)

15. By cancelling, S=√h×(h—a)×(h—b)×(h—c)

COMPUTATION OF THE CANON.

Having been somewhat full on the theory and construction of logarithms of numbers, it is intended to bestow considerable attention on the computation of the natural sines, tangents, &c. It is believed that a thorough knowledge of of the tables, and dexterity in using them, cannot well be attained without understanding thoroughly the method of computation. But as this subject is yet unprepared, it will be reserved for the appendix; nor, is it essential that it should be inserted here, as Day has placed it in a section by itself.

PARTICULAR SOLUTIONS OF TRIANGLES.

Reduction of the formula.

1. By Art. 232, AC: AB::1 cosA

2. By Alg. 388, 389, AC +AB AC-AB::1+cosA:

1-cosA

[blocks in formation]

6. By Art. 141, (Fig. 20.) BC2 =AC2 — AB2=(AC—AB)

×(AC+AB)

7. By Art. 232, tan2a=(AC-AB)÷(AC+AB)

8. Multiplying the 6th into the 7th, tanja XBC (AC — AB)X(AC-AB)=(AC-AB)2

9. Extracting the square root, tan a× BC=AC-AB 10. By Alg. 189, tan la 1::AC-AB: BC=Formula I.

11. Divid. the 7th by the 6th, tana÷BC2-(AC-AB) ÷(AC+AB)÷(AC−AB)×(AC+AB); therefore,

12.

tanla÷BC3=1÷÷(AC+AB)

13. Extract. the square root, tan ja÷BC=1÷(AC+AB) 14. Multiplying by tanļa, tanļa ×(AC+AB)2=BC 15. By Alg. 189, tanja 1::BC: AC+AB, or 1: tanja ::AC+AB: BC=Formula II.

16. By similar triangles (Fig. 20.) AB BC::1 tan a 17. By Alg. 389, AB+BC: AB-BC::1+tan a : 1-tan a 18. By Alg. 382, AB+BC: AB--BC: :(1+tan a)÷(1+ tan a) (1-tan a)÷(1+tan a); also, AB+BC: AB-BC::1 (1-tana)-(1+tana)

19. By Art. 116, the sum of the angles A and C-90°, and their sum =45°

2

20. In Art 218, let a=45° and B=A, then tan (a+b)= R2 (tan a-tan b)÷(R2+tan a tan b)

21. By substituting. tan(45°-A)=1(tan45°-tanA)÷(1+ tan45°+tan4), and as tan 45°=R=1(Art 95.) 22. By substituting, tan(45°-A)=(1-tanA)÷(1+tanA) 23. Subst. in 17th, AB+BC: AB-BC::1: tan(45°-A) 24. By inversion, 1 tan(45°-A)::AB+BC:ÀB–BĆ =Formula III.

Six examples under article 233.

Example 1. By Formula I, tan 14:1::AC-AB ; BC. By the conditions of the question, tan 1.4=tan (33°)=tan! (33° 45')=tan(16° 52′ 30′′)=9.4819392: Also, AC-AB 64, whose logarithm=1.8061800; therefore, by the formula, 9.4819392: 10.0000000::1.8061800;2.3242408=211 the perpendicular BC.

Eaxmple 2. By Formula II, 1: tanļa::AC+AB; BC. By the question, tan,a=tan! 37°=tan 18° 30′ 9.5245199. Again, AB + AC=1853, whose logarithm=3.2678754; therefore by the formula, 10.0000000: 9.5245199:: 3.2678754 2.7923953=620.

Exam. 3. Formula III, 1; tan (45°-A):: AB+BC: AB -BC; and, substituting numbers 1: tan 45° 41° 15′ : : 128.4 AB-BC. By the tables, 10.0000000 : 8.8165294 ::2.1085650; 0.9250944-8.416 the difference between

the base and perpendicular. By Alg. 341, į sum+¦ diff. the base 128.4+18.4-64.2+4.2=68.4; and, by Alg. 341, sum-diff.=the perpendicular=128.4-18.4-64.2 -4.2=60. Therefore, the base =68.4, and the perpendicular 60.

Example 4. By Formula II, 1: tan! C::AC+BC: AB. But the tan! C=tan! 40°-tan20°. By the tables, 10.0000000 :9.5610659::1.9190781: 1.4801440-30.21=the base.

Example 5. By Formula I, tanja;1::AC-BC: AB. But the tanla tan] 37° 15'-tan 18° 37′ 30′′, and ACBC=16.5: therefore, by the tables, 9.5276695; 10.0000 000::1.2174839; 1.6898144=48.96=the base.

Example 6. Formula III, by inversion, becomes tan (45° —A):1::AB-BC; AB+BC, and tan (45°-C):1:: AB-BC: AB+BC. But tan (45°-C)=tan(45°-27° 15')=tan 17° 45'. By the tables, 9.5052891: 10.0000000 ::1,5440680;2.0387699=109.3=the sum of the sides AB, BC. By Alg. 341, 54.65+17.5=72.15=the perpendicular, and 54.65-17.5=37.15=the base.

VARIOUS EXPLANATIONS UNDER ART. 234.

Theorem B. 2bc: a2 —(b2+c2)::R: cosA. In Figure 24, let AD=d, and make cot radius, then b;d::R: cosA, and by multiplying the first couplet by 2c, it becomes 2bc: 2dc R cosA. By Euc. II, 12, 2dc=a2-(b2+c2), and by substitution, we have the proportion 2bca2-(b2 +c2) R: COSA.

From Theorem B we proceed to derive expressions, both for the sides of a triangle, and for the cosines of the angles.

By the preceding, 2bc; b2+c2-a3::R cosA. Multiplying extremes and means, 2bc cosA=RX (ba+c2-a2): Dividing by 2bc, cosA=(b2+c2-a)÷2bc. Again, Ra2= R(b+c)-2bc cos a; and a2=b2+c2-(2bc cosA)÷R: Extracting the square root, a=√¿a +c2 − (2bc cosA)÷R)

Making BC radius, (Fig. 23.) and representing BD by d, a: d::R¦ cosA; multiplying by 2c, 2ac: 2dc: :R; cosA.

« ZurückWeiter »