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The solution of the cases of oblique plane triangles.,
ang Cop. and ABCC, and the sum subtracted from to one of 'em,
18ogives theother angleABC Two sides The other Let the angle ABC he found, AB, BC and side AC by the preceding case, and 3 an opp. an
then it will be, lin. C:AB:: gle C
tin. ABC: AC (by Theor.III.) Two sides The other As sum of AB and äc: their AC, AB and angles C dih: tang. of half the sum of
the included and ABC ABC and C:tang of half their 14 angle A
. (by Theor:V.) which added to, and subtracted from, the
half furn, gives the iwo angles. Two fides The other Let the angles be touiu oy 5 AC, AB and fide BC the last cal, and then BC the incl <A
by cafe 1. All the three An angle, Let fall a perp. BD) opp. to the fides. fuppose A req.angle: then (by Theor. IV.)
as AC:fum of AB and BC::
their dif. : dift. DG of the perp. 6
from the middle of the bale;
Note, The 2d 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) : fin. C. :: BC': sin. A (or of Cab,) may, it is evident, be either the acute angle A, or the obtuse one CaB (which is its supplement), the fines of both being exactly the same.
Having laid down, the method of resolving the different cases of piane 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 arch being given, to find its cofine, versed fine, tangent, fo-tangent, secant, and Co-fecant.
will be known. Then because of the similar triangles CFE, CAT, and CDH, it will be (by 14. 4.)
1. CF:FE::CA: CT; whence the tangent is known.
2. CF: CE (CA) :: CA: CT; whence the secant 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 secant 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 likewife 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. Q Xco-tang. Q; therefore will co
tang. P : čo-tang. Q: : tang. Q :tang. P; ot as co-tang. P: tang. Q:: co-tang. Q: tang. P (by 10. 4.)
2 PROP. II.
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 snean, to balf the sum of the fines BE+DG, of the two extremes : and, as radius to the fine of the commun difference, fo is the co-fine FO of the mean, to half the difference of the lines of the two extremes.
For let BD be drawn, interfecting the radius OC
in m; also draw В.
mn parallel to CF, meeting AO in ni
and BH and mv, AE, F!! 6 0
parallel to AO, 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 bisects it (by 1:3.) and therefore Bm for Dm) will be the fine of BC (or DC), and Om its co-sine: moreover mn, being an arithmetical mean between the sines BE, DG of the two extremes (because Bm = Dm) is therefore equal to half their sum, and Du equal to half their difference. But, because of the similar triangles OCF, Omn and Dum,
It will be SOC :0m:: CF: mn ?
Because of the foregoing proportions, we have
2 DmX FO
20m X CF and therefore DG+BES OC
OC. 2DmX FO. and DGBEE
Hence, if 'the mean arch AC 'be supposed that of 60°; then Of being the co-fine of 60°, ='sine 30°= 1 chord of 60" = TOC, it is manifest that DG-BE will, in this case, be barely=Dm; and consequently DG =Dm + BE. From whence, and the preceding corollary, we have these two useful theorems.
1. If the fine of the mean, of three equidifferent arches (supposing radius unity) be multiplied by twice the co-line of the common difference, and the line of either extreme be subtracted from the produet, the remainder will be the fine of the other extreme.
2. The fine of any arcb, above 60 degrees, is equal to the fine of another arch, as much below 60°, together with the fine of its excess above 60".
Om X CF
taken in a geometrical sense, die
OC notes a fourtb-proportional to OC, Om and CF; but, arithmetically, it signifies the quantity arising by dividing the produet of the measures of Om and CF by that of OC. Underfand tbe like of others.