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The solutions of the cases of right-angled plane trianglés.


the angles

Given Sought

The hyp: One leg As radius is to the line of
IAC and BC A, fo is the hyp. AC to

the leg BC (by Theor. I.) The hyp. The an- As AC : BC :: radius : 2 AC and gles fin. A (Theor. 1.). whose one leg BC complement is the angleC. The hyp. The other Let the angles

be found, 3 AC and leg AB by Cafe 2. and then the re

one leg BC quired leg AB, by Case 1.

The an- The hyp. As fine A : radius :: the 4 gles and AC leg BC:

to the hyp one leg BC AC (Theor. I.)

The an- The other As fine A: BC :: fine C

gles and leg AB : AB (by Theor. III.) Or, 5 one leg

as radius : tang. C :: BC BC

: AB (by Tbeor. II.) The two The an- As AB : BC :: radius : 6 legs AB

gles tang. A (by Theorem II.) and BC

whose complement is the

angle C. The two The hyp. Let the angles be found, 7 legs AB

AC by Cafe 6. and then the and BC

hyp. AC, by Cafe 4.





C D A तव The solution of the cases of oblique plane triangles.



Given Sought

Solution. The angles Either of As fine C: AB :: fine A. I and one fide the other BC (by Theor. Ill.).

AB sides BC Two sides The other As AB: fin.C:: BC : sin. A AB, BC and angles A (byTheor. III.) which added to an ang. Cop. and ABC|C, and the sum fubtracted from to one of'em.

180 gives the other angle ABC. Two sides The other Let the angle ABC be found,

AB, BC and fide AC by the preceding case, and 3 an opp. an

then it will be, sin. C:AB :: gle C

fin. ABC:AC (by Theor. Ill.) Two sides (The other As sum of AB and AC : their AC, AB and angles Cdif. :: tang. of half the sum of

the included and ABC ABC and C: tang. of half their 4 angle A


. (by Theor. V.) which added to, and subtracted from, the

half fum, gives the two angles. Two sides The other Let the angles be found by 5 AC, AB and fide BC the last case, and then BC, the incl.LA.

by Cafe 1. All the three An angle, Let fall perp.

BD sides suppose Afreq. angle: then (by Theor. IV.)

as AC: sum of AB and BC:

their dif.: dift. DG of the perp. 6

from the middle of the base; whence, AD being also known, the angle A will be found by Cafe 2. of right-angles.


opp. to the


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Note, The ed and 3d cases are ambiguous, or admit of two different answers each, when the side AB opposite the given angle C (see fig. 2.) is less than the given side BC, adjacent to it (except the angle found is exactly a right one): for then another right-line Ba, equal to BA, may be drawn from B to a point in the base, somewhere between C and the perpendicular BD, and therefore the angle found by the proportion AB (aB) : sin. C :: BC : fin. A (or of CaB,) may, it is evident, be either the acute angle A, or the obtuse one CaB (which is its fupplement), the fines of both being exactly the same

Having laid down the method of resolving the different cases of plane triangles, by a table of fines and tangents; I shall here shew the manner of constructing such a table (as the foundation upon which the whole doctrine is grounded); in order to which, it will be requisite to premise the following propositions.

The fine of an arcb being given, to find its co-
fine, versed fine, tangent, co-tangent, secant, and

IT Let AE be the proposed

arch, EF its fine, CF its D H

co-fine, AF its versed sine,

AT its tangent, CT its seE

cant, DH its co-tangent, and CH its co-fecant.

Then (by 8. 2.) we have CF = the fquare root of

CE? EF; whence, not "A only the co-fine CF, but also the versed fine AF,


will be known. Then because of the similar triangles CFE, CAT, and CDH, it will be (by 14. 4.)

1. CF:FE::CA: AT; whence the tangent is known,

2. CF : CE (CA) :: CA: CT ; whence the fecant is known.

3, EF : CF :: CD : DH; whence the co-tan

gent is known.

4. EF : EC (CD) : : CD : CH; whence the co-secant is known.

Hence it appears,

1. That the tangent is a fourth proportional to the co-fine, the sine, and radius.

2. That the fecant is a third-proportional to the co-fine and radius.

3. That the co-tangent is a fourth-proportional to the fine, co-fine, and radius.

4. And that the co-secant is a third-proportional to the fine and radius.

5. It appears moreover (because AT:AC::CD (AC): DH), that the rectangle of the tangent. and co-tangent is equal to the square of the radius (by 10. 4.): whence it likewise follows, that the tangent of half a right-angle is equal to the radius ; and that the co-tangents of any two different arches (represented by P and Q) are to one another, inversely as the tangents of the same arches : for, since tang. P x co-tang. P

squ. rad. = tang. Qx co-tang. Q; therefore will co


tang. P: co-tang. Q:: tang. Q: tang. P; or as co-tang, P: tang. Q:: co-tang. Q: tang. P (by IO. 4.)


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If there be three equidifferent arches AB, AC, AD, it will be, as radius is to the co-fine of their common difference BC, or CD, so is the fine CF, of the mean, to half the sum of the fines BE + DG, of the two extremes : and, as radius to the fine of the common difference, so is the co-fine FO of the mean, to half the difference of the fines of the two extremes, 1)

For let BD be drawn, interfect, ing the radius OC

in m; also draw

min parallel to CF,
meeting AO in n;

and BH and mu,
G 0

parallel : toAO meeting DG in H and v.

Then, the arches BC and CD being equal to each other (by hypothesis), OC is not only perpendicular to the chord BD, but also bifects it (by 1. 3-) and therefore B m (or Dm) will be the fine of BC (or DC), and Om its co-line: moreover 7 n., being an arithmetical mean between the fines BE, DG of the two extremes (because Bm = Dm) is therefore equal to half their fun, and Du equal to half their difference. But, because of the similar triangles OCF, Omn and Dum,


It will be SOC : Om :: CF : mn?
OC : Dm :: FO: Dv

}2. E. D.


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