Sidebilder
PDF
ePub
[ocr errors]

Pi. Trig,

PRO P. VII.
V any triangle, as the rectangle contained by two

sides, is to the rectangle contained by half the fum, and half the difference of the base and the difference of the fides, fo is the square of the radius to the square of the fine of half the angle opposite to the base.

Let ABC be a triangle, of which the side AC is greater than AB; and make AD equal to AC; therefore BD is the difference of the sides AC, AB: The rectangle BA, AC is to the rectangle contained by half the sum, and half the difference of CB, BD, as the square of the radius to the square of the fine of half the angle BAC.

Join CD, and draw a AE, BF perpendicular to CD, and from the centre B, at the distance BC, describe a circle, meeting AD in G, H, and CD in K: and because

AD is equal to AC, the angle ACD B 5.1. is equal to ADC, and AED, AEC

A are right angles, and the fide AE is

common to the triangles AED, AEC; € 26. 1. therefore DE is equal to EC, and

Б the angle DAE to EAC: and because

BF drawn through the centre cuts CK d 3. 3. at right angles, it bisects it d; therefore CK is double of CF: and CD

K is double of CE; therefore the remainder DK is double of the remain

der EF: and because the straight lines CK, GH in the circle, € 35. 3. cut one another, the rectangle CD, DK is equal e to the rect

angle GD, DH; and the rectangles contained by their halves are equal; therefore the rectangle DE, EF is equal to the rectangle contained by half of DG, the sum of CB, BD, and half of

DH their difference: but because BF is parallel to AE, DA is 12. 6. to AB, as DE to EF *; therefore the rectangles DA, AB, and 81.Def.6. DE, EF, are fimilar figures 8 upon AB, EF: and the squares of

AD, DE are similar; therefore the rectangle DA, AB, or CA, h 22. 6. AB, is the rectangle DE, EF, as the square of AD to the square

of DE ; that is, as the square of the radius to the square of the k s. P. T. fine of DAE \, half of BAC. Therefore the rectangle CA,

AB is to the rectangle contained by half the sum, and half the difference of CB, BD as the square of the radius to the square of the fine of half the angle BAC.

PROP.

[ocr errors]

Pl. Trig.

PROP. VIII.
N any triangle, the rectangle contained by two

sides, is to the rectangle contained by half the
sum of the three sides, and its excess above the base,
as the square of the radius to the square of the coline
of half the angle opposite to the base.

f 3• 3,

Let ABC be any triangle, the rectangle BA, AC is to the rectangle contained by half the sum of the three fides and its excess above the base BC, as the square of the radius to the {quare of the cofine of half the angle BAC.

In BA produced, take AD equal to AC; and join DC, and
draw a AE, BF perpendicular to DC; and from the centre B, a 12. 2. *
at the distance BC, describe a circle, cutting DC again in K,
and BD in H, G: and bisect 6 DG in L: and because AD is b 10. I.
equal to AC, the angles ADC, ACD are equal ©: and AED, .C gos
AEC are right angles; therefore DE is equal d.to EC: and d 26. 1.
the angle BAC is equal to ACD, ADC, it is therefore double e 32. f.
of ADC: and because BF, from the centre B,
is at right angles to CK, it bisects it f; there-

D
fore CK is double of CF; and CD is double of
CE; therefore the remainder DK is double of
the remainder EF: For the same reason, DH is
double of BL: and because DG, DC cut the

e
circle, the rectangle GD, DH is equal to the

& Cor. 36.
rectangle CD, DK; and therefore the rectangies 4
contained by their halves are equal; that is,
the rectangle DE, EF is equal to the rectangle
contained by DL half of DG the sum of the

C
three sides, and BL the excess of LG that half
fum above BG or BC the base : but because
AE is parallel to BF, DA is to AB, as DE to
EFh; therefore the rectangle DA, AB is fini. G

h 16.6. lar k to the rectangle DE, EF; and the squares

ki. De!.6. of DA, DE are fimilar ; therefore the rectangle DA, AB, or CA, AB, is to the rectangle DE, EF, as the square of DA to the square of DE'; that is, as the square of the radius to the 122. ú. {quare of the cofine of ADC, half of the angle BAC m. Where- m l. P. fore, as the rectangle contained by the fides BA, AC is to the rectangle contained by half the sum of the three fides, and its excess above the base BC, so is the square of the radius to the {quare of the cofine of half the angle BAG.

COR.

[ocr errors]

3.

B

[ocr errors][ocr errors]
[ocr errors]
[ocr errors]

Pl. Trig. Cor. Because the rectangle contained by half the sum of the

three fides, and its excess above the base, is to the rectangle

BA, AC, as the square of the coline of half BAC to the square 6. P. T. of the radius ?: and that the rectangle BA, AC is to the rect

angle contained by half the sum, and half the difference of the

bale and difference of the sides, as the square of the radius to 0 7. P. T. the square of the fine of half BAC °; therefore, by equality P, P 22. 6. the rectangle contained by half the sun of the three sides, and

its excess above the base, is to the rectangle contained by half the sum, and half the difference of the base and difference of the fides, as the square of the cofine of half BAC to the square

of its fine ; that is, as the square of the radius to the square 9 2. Cor. of the tangent 9 of half the angle BAC opposite to the 19. Def.

base.

SOLUTION of the Cases of OBLIQUE-ANGLED

TRIANGLES.

GENERAL PROPOSITION.

N any

IN

triangle, of the three fides, and any two of the angles, any three being given, the other two and the remaining angle may be found.

But if the three angles be given, the ratios only of the sides are given, beir.g the same with those of the angles opposite to them; and in this case, the sides cannot be found.

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

If two angles of a triangle be given, the third is also given, being the supplement of their sum : and, on the contrary, if one of the angles be given, the sum of the other two is also given.

The

The four cases of this proposition may be resolved by the help. Pl. Trig. of some of the preceding propofitions ; as in the following Table.

I

:}P.3.

Caies. Given. sought.

solution.
A, two BC Sin. G:fin. A::AB:BC.

angles and a side. AC Sin. C:sin. B ::BA: AC.}
AB, AC and B, JAC: AB :: fin. B: fin. C. P. 3.

two sides, and an The If AC be greater than AB, C
angle oppofite angles. is acute: otherwise it may be
to one of them.

acute or obtuse, by Cor. to
Def. 8.

2

3

4

AB, AC and A, The The sum of BA, AC: diff. of BA,

two fides and angles AC :: tan. of į sum of B, C :
the included B & C tan. į diff. B, C. of P. 4.
angle. and BC. Otherwise, AB : AC::R: tan.

E; and R: tan. diff. of E and
45°: : tan. į sum of B, C : tan.

diff. of B, C. P. 6.
Whence B and Care found, by

LEM. 4.

Sin. B : fin. A::AC: BC.
AB, AC, BC, the A,B,C, Let AD be perp. to BC. BC:
three sides. the sum of BA, AC: diff. of BA,

three AC: F; and BC, F together
angles.

are double of the greater seg

ment BD. P.s.
If BD be greater than CB, C is

obtuse; if not, it is acute:

then AC: CD::R: cos. C.
Otherwise, Let D be the diff. of

AB, AC, the rect. AB, AC:
rect. fum BC, D and diff.

BC, D :: R2: fin 2. 1 BAC.
Otherwise, Let P be fum of the

three sides. The rect. BA,
AC: rect. contained by P and
the diff. of P, BC::R 2: cos 2.

BAC.

[blocks in formation]

k 1. 2.

Pl. Trig. To construct a Table of Sines, Tangents, &c.

Let ABC be a circle, of which the diameter is AC, and

the centre D: and let AB be any arch of it; and bisect AB in a 30. 3. Fa; and join AB, BC, BD, DF: and let DF meet AB in E; 0 120 l. and draw 6 BG purpendicular to AC: and because the arch

AB is double of AF, the angle ADB is C 33.6. double of ADFC: but ADB is double

B d 20. 3. of ACB, therefore the angle ADF is

equal to ACB; and DE is parallel to e 31.3. BC, or at right angles to AB e; theref 4. 6. fore AC is to CB, as AD to DEf; and

E
AC is double of AD; therefore CB is

C

D G A g 14. 5. double of DE 8: and because the tri.

angles ADE, BCG are equiangular, AD is to DE, as BC to

CG'; therefore the rectangle AD, CG is equal to the rectangle h 16.6. BC, DE “, that is, to double of the square of DE k: and DE is

the cofine of AF!, and CG the sum of the radius CD, and DG 1 1. Cor. the cofine of AB. Therefore, the rectangle contained by the Def, 10.

radius, and the sum of the radius and the cofine of Pl.T.

arch,

any is double of the square of the cofine of half that arch.

If AB be equal to the radius AD, it is the side of a hexagon m Cor. 15. inscribed in a circle ; and the arch AB is the fixth part of the

circumference; that is, it is an arch of 60°; and because then n 47. 1. the triangle ABD is equilateral, AG is equal to GD"; that is,

it is the half of the radius AD. If, therefore, the radius be represented by unity, and the decimal notation be used, the sum of the radius and the cofine DG is represented by 1.5, and the rectangle contained by the radius and this sum is also represented by 1.5, because the radius is 1 ; therefore half of this rectangle is .75, which is therefore the square of DE the cofine of AF; that is, of 30°. Wherefore the cofine of 30° is the square-root of .75; that is, .8660254039344+.

In the same manner, if the radius 1 be added to this number, half the sum .9330127019672+ is the square of the cofine of 15°; and therefore its square-root .96592582632789 + is the cosine of 15°. In like manner, the square-root of half the sum of i and the cofine of 15° is .99144485936634 + the cofine of go 30'. In the same manner are found the cosine of 3° 45', of 1° 57' 30', and so on : till at length, after twelve bisections of the arch of 60°, the cofine of 52" 44" 03"" 4.5"' is found to be -9999999673176965713. And if the square of the cosine be subtracted from i, which is the square of the radius; the remainder is the square of the fine of the same arch”. Thus, the

fine

4.

« ForrigeFortsett »